Exceptional Polynomials and Monodromy Groups in Positive Characteristic

Exceptional polynomials and monodromy groups in positive characteristic

Dissertation zur Erlangung des naturwissenschaftlichen Doktorgrades der Julius-Maximilians-Universit¨at W¨urzburg

vorgelegt von

Florian M¨oller

Institut fürMathematik der UniversitätWürzburg

Würzburg, März 2009 ii Danksagung Ich möchte mich herzlich bei Professor Peter Müllerfür seine Unterstützung in allen Phasen meiner Promotion bedanken. Als Betreuer meiner Dissertation lastete auf Herrn Müller die Aufgabe, mir einerseits zu zeigen wie schönfreie und eigenständige Forschungstätigkeit sein kann, mir andererseits aber auch Probleme vorzulegen, die mich in meiner Promotion voranbrachten. Dies ist Herrn Müller ausgezeichnet gelungen. Besonders erwähnen möchte ich noch, dass Professor Müller mehrere komplizierte math- ematische Situationen, mit denen ich mich konfrontiert sah, stark vereinfachte. Erst durch seine Hilfe sind mir einige Beweise gelungen oder in ihre jetzige lesbare(re) Form gebracht worden.

Weiterhin danke ich Professor Theo Grundhöfer,der für meine Fragen immer ein offenes Ohr hatte und mir an mancher Stelle mit wertvollen Literaturhinweisen half. Weiter habe ich ihm eine zusätzliche Anstellung am Lehrstuhl fürMathematik III zu verdanken. Ohne die Unterstützung von Herrn Grundhöfer wäre mir meine Promotion erheblich schwerer gefallen.

Mein besonderer Dank gilt meiner Familie, die mich in meinem Vorhaben stets best¨arkte und mir einen Ausgleich zur mathematischen Besch¨aftigung bot.

iii iv Contents

1. Introduction 1

I. General Results 5

2. Monodromy groups and affine polynomials 7 2.1. Polynomials and monodromy groups ...... 7 2.2. Affine groups and affine polynomials ...... 8 2.3. Generic polynomials ...... 10

3. Calculation of differents and the genus-0 condition 13 3.1. The general case ...... 13 3.2. Applications to the case of affine Galois groups ...... 18 3.2.1. Consequences of Hasse-Arf and Herbrand ...... 18 3.2.2. Bounds for the number of fixed points for elements of G ...... 20 3.2.3. Bounds for ind(∞)...... 20 3.2.4. Bounds for ind(p) with p a finite place ...... 21

II. Application to speciﬁc problems 23

4. Aﬃne polynomials with g ≤ 2 25 4.1. g =0...... 25 ∼ 4.1.1. Case (A): H = Cm is cyclic ...... 26 4.1.2. Case (B): H is an elementary abelian p-group ...... 28 ∼ 4.1.3. Case (C): H = (Cp × · · · × Cp) o Cm is a semidirect product . . . . . 28 4.1.4. Case (D): H is dihedral in even characteristic ...... 28 4.1.5. Case (E): H is dihedral in odd characteristic ...... 28 ∼ 4.1.6. Case (F): H = A5 in characteristic 3 ...... 28 4.1.7. Case (G): H =∼ PSL(2, q) with p 6=2 ...... 29 4.1.8. Case (H): H =∼ PGL(2, q) with p 6=2...... 37 4.1.9. Case (I): H =∼ PGL(2, 2m)...... 40 4.2. g =1...... 41 4.2.1. Ramiﬁcation in E|K(t) ...... 41 4.2.2. Cases in odd characteristic p > 2...... 43 4.2.3. Cases in even characteristic p =2...... 43 4.3. g =2...... 47 4.3.1. Cases in odd characteristic p > 2...... 47 4.3.2. Cases in even characteristic p =2...... 51

5. AGL as a monodromy group 53

v Contents

6. Aﬃne polynomials of degree p2 61 6.1. Cases in odd characteristic p 6=2...... 62 6.2. Cases in even characteristic p =2...... 63

7. Exceptional polynomials of degree p3 65 7.1. The cases in characteristic p > 3 ...... 65 7.2. The cases in characteristic p ∈ {2, 3} ...... 68

8. Exceptional polynomials of degree pr, r an odd prime, with 2-transitive group A 69

vi 1. Introduction

In 1923 Schur [43] requested a description of all polynomials f ∈ Z[X] that induce a bijection on Z/pZ for infinitely many primes p. He proved that if f is of prime degree, then it is – up to q linear changes over the algebraic closure of Q – either a cyclic polynomial X or a Chebychev q 1 polynomial Tq(X) (defined implicitly by Tq(X + 1/X) = X + Xq ). He conjectured that in the general case f is a composition of such polynomials. This was proved by Fried [13] in 1970. Schur’s original question has been generalized in several ways: Turnwald [48] discusses the problem over integral domains, Guralnick, Müller, and Saxl [19] characterize rational functions r over number fields K so that r induces a bijection on the residue field Kp for infinitely many places p ∈ P(K).

There is a different generalization of interest to us: we assume the base field to be of positive characteristic and search for exceptional polynomials, i.e. polynomials that fulfill the following

Definition (Exceptionality) Assume k is a finite field. Let f ∈ k[X] be a polynomial with coefficients in k. If K is an extension field of k, then f is called a permutation polynomial over K if f induces a bijection on K. f ∈ k[X] is called exceptional over k if it is a permutation polynomial over infinitely many finite extensions of k.

The classification of permutation polynomials has a long tradition and goes back to Her- mite [24] who noticed that any function k → k with k a finite field can be represented by a polynomial. A broad survey of permutation polynomials can be found in Lidl and Niederreiter [35]. An important step forward concerning the theory of exceptional polynomials was the refor- mulation of the original definition in terms of Galois theory and covering theory in positive characteristic. A first consequence was the proof of the following

Exceptionality Lemma (cf. [14]) Let k be a ﬁnite ﬁeld of characteristic p and t a transcendental over k. Let f ∈ k[X] be a polynomial. Denote the set of zeros of f(X) − t by Z. Fix a zero x ∈ Z. Then:

(1) Suppose f is not a p-th power in k[X]. Then f is exceptional over k if and only if the x-stabilizer of the arithmetic monodromy group of f ﬁxes no orbit of the x-stabilizer of the geometric monodromy group of f on Z \{x}. (A deﬁnition of monodromy groups is given on page 7.)

(2) Suppose f is a composition f = f1 ◦ f2 with f1, f2 ∈ k[X]. Then f is exceptional over k if and only if both f1 and f2 are exceptional over k. In 1993 Fried, Guralnick, and Saxl [14] realized that this result reduces the classiﬁcation of exceptional polynomials essentially to a question about primitive groups. They used the

1 1. Introduction

Theorem of O’Nan-Scott and the classiﬁcation of ﬁnite simple groups to obtain

Theorem Let k be a ﬁnite ﬁeld of characteristic p. Assume f ∈ k[X] is an indecomposable and exceptional polynomial of degree n. Denote the geometric monodromy group of f by G. Then one of the following holds:

(1) n is an odd prime diﬀerent from the characteristic p. The group G is cyclic or dihedral of degree n.

r r r (2) n = p and G = Fp o H is an aﬃne group with H ≤ GL(r, p) acting naturally on Fp. (3) p ∈ {2, 3}, there exists an odd integer a ≥ 3 with PSL(2, pa) ≤ G ≤ PΓL(2, pa), and 1 a a n = 2 p (p − 1). This classiﬁcation solved in particular Carlitz’s conjecture (1966): if p is an odd prime, then the degree of f is odd.

The three cases of the above theorem are understood with different degrees of completeness: Case (1) is classical; up to linear changes only cyclic or Chebychev polynomials of degree n arise here, cf. [38, Appendix]. This is the equivalent to Schur’s original conjecture. The first examples in case (3) were given by Müller [39] for p = 2 and a = 3. Cohen and Matthews [10] generalized these to an infinite series in even characteristic. Lenstra and Zieve [34] found a similar series for p = 3. Recently, Guralnick, Zieve, and Rosenberg [22, 20] gave a complete discussion of case (3). The polynomials occurring fulfill either G = PSL(2, pa) or G = PGL(2, pa). Case (2) is still open. The main problem is the difficulty to find restrictions for the group H; even the smallest possible case deg f = p requires some work, cf. [14, §5]. For some time the only examples of this case were additive polynomials and certain twists of them. In 1997 Guralnick and Müller [17] found a completely new series. Guralnick and Zieve [22] conjecture that there are no more polynomials belonging to this case. Up to the present every example fulfills the following

Observation. The fixed field E of the affine kernel of the geometric monodromy group of f is rational.

This observation motivates a classification of exceptional polynomials with primitive affine arithmetic monodromy group in terms of the genus g of E. This is done in chapter 4 up to g = 2. As a result, E is rational if and only if f belongs to a family of known polynomials. Moreover there are no exceptional polynomials with primitive affine arithmetic monodromy group for g ∈ {1, 2}. However, Theorem 4.25 shows that in case g = 2 the affine group AGL(2, 3) can be realized as the geometric monodromy group of a polynomial. Chapter 5 generalizes this AGL(2, 3)-polynomial; Theorem 5.7 eventually shows that every affine group AGL(r, pe) can be realized as the geometric monodromy group of a polynomial re of degree p . Unfortunately, these groups are 2-transitive on the r-dimensional Fpe -vector space; hence, this chapter does not offer exceptional polynomials at all. The last three chapters continue the classification of [14, §5]. In chapter 6 polynomials of degree p2 with primitive and affine arithmetic monodromy group are discussed. Theorem 6.1 shows that such a polynomial either belongs to a class of known affine polynomials or has a “big” geometric monodromy group. Chapter 7 classifies exceptional polynomials of degree

2 p3 with primitive arithmetic monodromy group. Again only known exceptional polynomials are obtained. In chapter 8 we study exceptional polynomials of degree pr where r is an odd prime. We assume additionally that the arithmetic monodromy group of f is 2-transitive. Theorem 8.1 shows that this chapter does not oﬀer new classes of exceptional polynomials.

Notation and terminology Mostly we use standard notation. In the following we give terminology about which an explanation may be needed. For an integer m ∈ N we denote by Cm resp. Dm a cyclic group of order m resp. a dihedral group of order 2m. Assume that E is a function ﬁeld. Then g(E) is the genus of E. The set of places of E is denoted by P(E). Page 23 gives a survey of frequently used deﬁnitions.

3 1. Introduction

4 Part I.

General Results

2. Monodromy groups and aﬃne polynomials

In this chapter we give the basic ideas how Galois theory and ramiﬁcation theory can be used to translate properties of polynomials into properties of certain ﬁeld extensions. Most of the following results are classical.

2.1. Polynomials and monodromy groups

Monodromy groups Let k denote a finite field of characteristic p, and let f ∈ k[X] \ k[Xp] be a polynomial of degree n which is not a p-th power in k[X]. Suppose t is transcendental over k. Denote by K an algebraic closure of k. Set ` the splitting field of f(X) − t|k(t). As f(X) − t ∈ k(t)[X] is separable, the extension `|k(t) is Galois. Its Galois group A := Gal`|k(t) is called the arithmetic monodromy group of f. Let k0 ⊆ K denote an algebraic extension of k; obviously t is also transcendental over k0. The splitting field of f(X) − t|k0(t) coincides with the compositum k0`. Set M := Galk0`|k0(t).

The restriction of any Galois automorphism σ ∈ M to ` yields a Galois automorphism σ|` ∈ A. Since the mapping M → A, σ 7→ σ|` is injective, we obtain an embedding of M into A; hence, we can consider M to be a subgroup of A. Denote the exact ﬁeld of constants of ` by k˜. As k(t) ⊆ ` ∩ k0(t) ⊆ k0(t), L¨uroth shows that ` ∩ k0(t) = (k˜ ∩ k0)(t). Since the extension (k˜ ∩ k0)(t)|k(t) is Galois with Galois group Gal(k˜ ∩ k0)(t)|k(t) =∼ Gal(k˜ ∩ k0|k), it follows in particular that M is a normal subgroup of A with A/M being cyclic of order [(k˜ ∩ k0): k].

All in all, we have proved the following fact: For every choice of k0 the group M contains the group G := Gal`|k˜(t) as a normal subgroup with M/G being cyclic. G is called the geometric monodromy group of f. Note that M is isomorphic to G in particular if k0 = K. As we work later on mostly with function ﬁelds whose ﬁelds of constants are algebraically closed, this easy consequence becomes important.

Since f(X)−t is absolutely irreducible, the groups G, M, and A act transitively on the zeros of f(X) − t. Functional decomposability Deﬁnition 2.1 With the above notation we call f functionally decomposable over k0 if there exist nonlinear polynomials g, h ∈ k0[X] with f = g ◦ h. If f is not functionally decomposable over k0, we call f functionally indecomposable over k0.

7 2. Monodromy groups and aﬃne polynomials

Let x ∈ ` be a zero of f(X) − t. We state some consequences of the functional indecomposability of f over k0.

Lemma 2.2 The following statements are equivalent:

(1) f is functionally indecomposable over k0.

(2) The monodromy group M acts primitively on the zeros of f(X) − t.

(3) There is no proper intermediate ﬁeld between k0(t) and k0(x).

Proof. We show first that (2) and (3) are equivalent: By Galois duality the field k0(x) is the fixed field of the stabilizer Mx of x. Both the primitivity of M and the non-existence of 0 0 proper intermediate fields between k (t) and k (x) translate into the maximality of Mx in M. Now, suppose f = g ◦ h to be decomposed over k0. Let Z := xM denote the set of zeros of −1 f(X) − t. Set Y := h(Z) and define ∆y := h (y) ⊆ Z for y ∈ Y . It is easy to verify that {∆y | y ∈ Y } is a system of imprimitivity for M. Next, assume there is a proper intermediate field between k0(t) and k0(x). As this field is rational by Lüroth, there exists y ∈ k0(x) with k0(t) ⊂ k0(y) ⊂ k0(x). Since x resp. y is algebraic over k0(y) resp. k0(t), we can find nonlinear polynomials g, h ∈ k0[X] such that h(x) = y and g(y) = t. Thus, x is a zero of (g◦h)(X)−t. Consider f(X)−t and (g◦h)(X)−t as polynomials in t. Then both are irreducible and have the same leading coefficient. This gives the equality f = g ◦ h.

2.2. Aﬃne groups and aﬃne polynomials

Definition 2.3 Let G be a permutation group on a finite nonempty set Ω. We call G an affine group if G contains a normal subgroup N that fulfills the following two conditions:

• N is elementary abelian.

2 • N is regular on Ω, i.e. for every (ωi, ωj) ∈ Ω there exists exactly one n ∈ N with n ωi = ωj.

Remark 2.4 The elementary abelian regular normal subgroup in the above definition is not unique in general. For instance, in the affine group AGL(2, 3) there exists an affine subgroup of order 27 containing three regular elementary abelian normal subgroups. ?

We state some basic facts about aﬃne groups:

Lemma 2.5 Let G be an aﬃne group with N E G being elementary abelian and regular. Then:

(1) A point stabilizer H of G is a complement of N, G = N o H.

(2) There exist a prime p and an integer r such that N is isomorphic to the r-dimensional r Fp-vector space Fp.

8 2.2. Aﬃne groups and aﬃne polynomials

(3) For g ∈ G write g = nghg with ng ∈ N and hg ∈ H. Then the action of G on Ω and of G on N deﬁned by g −1 n := hg nnghg are equivalent. In particular, G can be embedded into AGL(r, p), the group of all aﬃne r transformations of Fp. H acts on N as a subgroup of GL(r, p). Proof. (1) is a direct consequence of the transitivity of N, cf. [32, 3.1.4]. (2) is merely another formulation of N being elementary abelian. (3) follows from Huppert [25, II 2.2].

The next deﬁnition connects aﬃne groups with polynomials. It is central for our further considerations:

Definition 2.6 With the above notation we call f ∈ k[X] \ k[Xp] an affine polynomial if its geometric monodromy group is an affine group. A consequence of Dixon and Mortimer [12, Sec. 4.7] and Huppert [25, II 3.2] is

Deﬁnition/Lemma 2.7 With the above notation G is primitive if and only if H acts irreducibly on N. If G is primitive, then N is the unique minimal normal subgroup of G. In this case we call N the aﬃne kernel of G. Huppert [25, II 3.2] also gives the important

Proposition 2.8 (Galois) Suppose G is a primitive affine group whose affine kernel has order pr. Then a point stabilizer H of G does not contain a nontrivial normal p-subgroup. As a first application we classify all affine polynomials having a regular geometric monodromy group.

Proposition 2.9 Let K be an algebraic closure of the field Fp. Denote by t a transcendental element over K. (1) Assume f is a semi-additive polynomial of degree n = pr, i.e. r X pi f(X) = a + aiX with a, ai ∈ K and a0ar 6= 0. (2.1) i=0 Then the geometric monodromy group G of f is elementary abelian and regular. In particular, f is affine. (2) Suppose f is affine and the geometric monodromy group G of f is a p-group. Then f is semi-additive of the form (2.1). Proof.

(1) Set g := f −a the linearization of f. Denote the set of zeros of g by Z := {z1, . . . , zn} ⊆ K and ﬁx a zero x of f(X) − t. Since g(zi + zj) = g(zi) + g(zj) = 0, Z is an Fp-vector space. The set of zeros of f(X) − t is given by x + Z; in particular, K(x) is the splitting ﬁeld of f(X) − t and |G| = [K(x): K(t)] = deg f = |Z|. It follows together with the transitivity of G that for every z ∈ Z there exists a unique gz gz ∈ G with x = x + z. The map gz 7→ z gives an isomorphism between G and Z. This is the claim.

9 2. Monodromy groups and aﬃne polynomials

(2) Let x denote a zero of f(X) − t and set H := Gx the stabilizer of x. As G is a p- ∼ group, there exist subgroups Gi with H = G0 E G1 E ··· E Gr = G and Gi+1/Gi = Cp. By Lemma 2.2 f is a composition of degree-p polynomials fi with monodromy group ∼ Gfi = Cp.

As Gfi is abelian of order p, it is regular. Turnwald [49, Theorem 2.10] proves that p there exist elements a, b, c ∈ K such that fi = aX + bX + c. A simple induction shows that f being a composition of semi-additive polynomials is semi-additive, too.

Remark 2.10 Part (2) of the above proposition can also be proved by using Corollary 3.16. We see that only the infinite place of K(t) ramifies in the extension K(x)|K(t); its ramification index equals [K(x): K(t)]. The fixed field of Gi is rational by Lüroth; thus, Fix(Gi) = K(yi) where yi can be chosen such that the infinite place of K(yi) lies over the infinite place of K(t). Hence, the extensions K(yi)|K(yi+1) are of degree p and without finite ramification. This p shows that yi fulfills an equation of the form ayi + byi + c = yi+1 with a, b, c ∈ K. Therefore the polynomials fi are semi-additive of degree p. ?

2.3. Generic polynomials

In this section we give an alternative approach to aﬃne polynomials. Our main idea is to ﬁnd a primitive element z for the extension k(x)|k(t) that has a simple minimal polynomial. Our main tool will be the following result of Kemper and Mattig [31]:

Proposition 2.11 Let k be an algebraic extension of Fp. Suppose t1, . . . , tr+1 are algebraically independent transcendentals over k. Then

r pr X pr−i g(t1, . . . , tr+1; X) := X + tiX + tr+1 ∈ k(t1, . . . , tr+1)[X] (2.2) i=1 is generic for AGL(r, p) over k, i.e. g fulﬁlls the following two conditions:

(1) The Galois group of g (as a polynomial in X) is AGL(r, p).

(2) If K is an infinite field containing k and L|K is a Galois field extension with Galois group H ≤ AGL(r, p), then there exist λ1, . . . , λr+1 ∈ K such that L is the splitting field of g(λ1, . . . , λr+1; X) over K.

The next lemma describes how the Galois group of g acts on the zeros of g.

Lemma 2.12 Let g be defined as in (2.2). Denote by L the splitting field and by G =∼ AGL(r, p) the Galois group of g. Fix a zero z of g. Then the set Z of zeros of g is given by Z = z + V where V ⊂ L is an Fp-vector space. Elements of the stabilizer Gz of z act on V as automorphisms of V . Elements of the affine kernel N of G stabilize V pointwise and act on Z by translation.

10 2.3. Generic polynomials

Proof. V is given as the set of zeros of the linearization of g, i.e. as the set of zeros of g(t1, . . . , tr, 0; X). As this polynomial is additive, it follows at once that V is an Fp-vector space. We prove that G acts on V . Let σ ∈ G. As zσ ∈ Z, there exists v0 ∈ V with zσ = z + v0. This gives for any v ∈ V (z + v)σ = z + v0 + vσ ∈ Z ⇐⇒ v0 + vσ ∈ V ⇐⇒ vσ ∈ V. Since we have σ σ σ σ σ (λv1) = λv1 and (v1 + v2) = v1 + v2 for all λ ∈ Fp and v1, v2 ∈ V, G acts on V as a subgroup of GL(V ) =∼ GL(r, p). Set ϕ : G → GL(V ) the homomorphism that maps σ to the V -automorphism induced ∼ by σ. Then ker ϕ ∩ Gz = 1. Since Gz = GL(r, p), the mapping ϕ is an epimorphism with | ker ϕ| = pr. As AGL(r, p) is primitive and, hence, N is the unique normal subgroup of order r p , N = G(V ) is the pointwise stabilizer of V . The remaining assertions follow.

Remark 2.13 Let t be a transcendental over k. Consider the polynomial

h(X) := g(a1, . . . , ar+1; X) ∈ k(t)[X]. (2.3) h results from g by specializing elements ai ∈ k(t) for the transcendentals ti. Suppose h is separable. Then the Galois group of h is a subgroup of the Galois group of g, cf. [33, VII §2]. Hence, our results from the above lemma can be transferred to describe the action of the Galois group of h. ?

Remark 2.14 Note that the Galois group of a polynomial of the form (2.3) is merely a subgroup of AGL(r, p); it need not be an aﬃne group. For instance, consider the polynomial 8 2 2 f(X) := X + tX + tX + t ∈ F2(t)[X]. Huppert [25, p. 161] proves that AGL(3, 2) contains a transitive complement C of its aﬃne kernel. We show that the Galois group G := Gal f|F2(t) of f is conjugate to C.

The action of G on the zeros x1, . . . , x8 of f gives a natural embedding of G into S8, the symmetric group on the set Ω := {1,..., 8}. Specialization of elements of F8 for the parameter t and factorization of the resulting polynomial over F8 show that G contains permutations of type [1, 1, 3, 3], [1, 7], and [4, 4]. A survey of the subgroups of AGL(3, 2) proves that this condition enforces either G = AGL(3, 2) or G = Cg for some g ∈ AGL(3, 2). Define Ω4 := {M ⊆ Ω | |M| = 4} to be the set of all subsets of Ω of cardinality 4. The g g definition M := {m | m ∈ M} for g ∈ G and m ∈ M gives an action of G on Ω4. If G = AGL(3, 2), then Ω4 splits into two G-orbits; in the other case Ω4 splits into three G-orbits. The MAPLE-function ‘galois/rsetpol‘ allows us to compute the polynomial Y Y f4(X) := X − xi ∈ F2(t)[X] M∈Ω4 i∈M whose zeros are the products of four pairwise different zeros of f. As f4 has three irreducible factors over F2(t), the claim follows. ?

11 2. Monodromy groups and aﬃne polynomials

Lemma 2.15 Let p be a prime and r ∈ N be an integer. Let E1,E2 ≤ Spr be elementary g abelian regular subgroups of Spr . Then there exists an element g ∈ Spr such that E1 = E2. In particular, all subgroups of Spr that are isomorphic to AGL(r, p) are conjugate.

Proof. The action of E1 resp. E2 is equivalent to the natural action of E1 on the coset ∼ space E1/1 resp. of E2 on the coset space E2/1. As E1/1 = E2/1, these actions are equivalent, too. Hence, E1 acts equivalently to E2. This shows that E1 and E2 are conjugate. 0 ∼ 0 ∼ Let A, A ≤ Spr with A = A = AGL(r, p). Denote the aﬃne kernel of A with N and 0 0 0 0 0 the aﬃne kernel of A with N . Then A = NSpr (N) and A = NSpr (N ). As N and N are conjugate, the same holds for A and A0.

Now we prove the main result of this section.

Proposition 2.16 Let k be an algebraic extension of Fp, denote by t a transcendental over k, and let g be deﬁned as in (2.2). Suppose f ∈ k[X] \ k[Xp] is an aﬃne polynomial of degree n = pr. Fix a zero x of f(X) − t. Then there exists an element z ∈ k(x) such that k(t, z) = k(x) and the minimal polynomial µ of z over k(t) is separable with µ(X) = g a1, . . . , ar+1; X and ai ∈ k[t]. (2.4)

Proof. Identify the zeros of f(X)−t with integers 1, . . . , n and set G ≤ Sn the permutation representation of Galf(X)−t|k(t) on the zeros of f(X)−t. Let N ≤ G denote an elementary ∼ abelian regular normal subgroup of G. Then A := NSn (N) = AGL(r, p), the aﬃne kernel of A coincides with N, and for every integer i ∈ {1, . . . , n} the stabilizer Gi of i fulﬁlls

Gi = Ai ∩ G.

The previous lemma shows that there exists an identification of the zeros of g with the integers 1, . . . , n such that the action of the Galois group of g on the zeros of g is given by A. Due to Proposition 2.11 and the proof of Kemper [30, Theorem 1] we can find elements ai ∈ k(t) such that the polynomial h(X) := g(a1, . . . , ar+1; X) is separable, the splitting field of h over k(t) coincides with the splitting field of f(X) − t over k(t), and the action of the Galois group of h on the zeros of h is equivalent to the action of G. Hence, we can find a zero z0 ∈ k(x) of h with k(x) = k(t, z0). 0 Set d the least common multiple of the denominators of the ai. Define z := dz . Then k(t, z0) = k(t, z) and a simple calculation shows that the minimal polynomial of dz over k(t) is of the form (2.4).

Example 2.17 Use the notation from the above proof. Suppose f is a sublinearized polynomial of the form (4.2), cf. page 28. Then we can write

X pi−1 m m f(X) = X · g (X) with g(X) = g pi−1 X ∈ k[X]. m m|pi−1 pi|n

Some calculation shows that we can set z0 := g(x)−1. Suppose f fulﬁlls the conclusion of Guralnick/M¨uller [17, Theorem 1.4]. Then we can set 0 1 z := f 0(x) . This follows from [17, §3]. ?

12 3. Calculation of diﬀerents and the genus-0 condition

In this chapter we present several techniques to calculate the diﬀerent exponent of an extension of a place. For the convenience of the reader we ﬁrst state some classical results that will help us with the following computations.

3.1. The general case

The first theorem describes how the different exponent d(P|p) of an extension P|p is related to its ramification groups. Since these groups are only defined in Galois extensions, we have to assume the field extension E|F to be Galois. A proof of the theorem can be found in Stichtenoth [47, III.8.8].

Theorem 3.1 (Hilbert’s Different Formula) Let L|E be a finite Galois extension of function fields, p ∈ P(E), and P ∈ P(L) lying over p. Let Ip(i) denote the i-th ramification group of P|p. Then the different exponent d(P|p) is given by

∞ X d(P|p) = |Ip(i)| − 1 . i=0

Next, we state a “transitivity formula” for diﬀerent exponents, cf. [47, III.4.11].

Lemma 3.2 Let E ⊆ F ⊆ L be a tower of separable extensions of function ﬁelds. Suppose p00 resp. p0 resp. p is a place of L resp. F resp. E with p00|p0 and p0|p. If e(p00|p0) denotes the ramiﬁcation index of p00|p0, then

d(p00|p) = e(p00|p0) · d(p0|p) + d(p00|p0).

The following theorem gives a lower bound for the diﬀerent exponent; moreover, tame extensions are characterized. Again, a proof can be found in Stichtenoth [47].

d(P|p) = e(P|p) − 1 ⇐⇒ P|p is tame.

We are still missing a comfortable tool to compute the degree of the diﬀerent in non-Galois extensions. It seems that the following proposition is widely known and often used, but the author does not know any reference for it. Hence, we will prove

13 3. Calculation of diﬀerents and the genus-0 condition

Proposition 3.4 Let L|E be a Galois extension of function fields with Galois group G. Let H ≤ G be a subgroup of G and set F := Fix(H) the fixed field of H. Define n := [G : H]. Consider G as a permutation group on n points via the natural action of G on the left-coset space G/H. Denote by o(S) the number of orbits of a subgroup S ≤ G. Let p ∈ P(E) be a place of E. Fix an arbitrary place π ∈ P(L) lying over p, denote by I(i) the i-th ramification group of the extension π|p, and define

∞ X n − o I(i) ind(p) := deg(p) . [I : I(i)] i=0 Then X ind(p) = d(q|p) deg(q) ∈ N0 q|p, q∈P(F ) equals the degree of the different in the extension F |E coming from the ramification of p. In P particular, p∈P(E) ind(p) equals the degree of the different of the extension F |E.

Proof. For a place q ∈ P(F ) denote by nq the number of places of L lying over q. If P ∈ P(L) is a place of L, denote by PF := P ∩ F the well-deﬁned restriction of P to F and by IP|PF (i) the i-th ramiﬁcation group of the extension P|PF . If q resp. P appears as an index of summation, then the sum extends over all places of F resp. L.

We use the notation from Stichtenoth [47]. We get

X (a) X 1 d(q|p) deg(q) = d(PF |p) deg(PF ) nP q|p P|p F

(b) X e(P|PF )f(P|PF ) = d(P |p) deg(P ) |H| F F P|p deg(π) X = e(P|P )d(P |p) |H| F F P|p (c) deg(π) X = d(P|p) − d(P|P ) |H| F P|p (d) deg(π) X X = |I(i)| − |I (i)| |H| P|PF P|p i≥0 (e) deg(π) X X = |I(i)| − |I (i)| |H| P|PF i≥0 P|p (f) deg(π) X |G| X = |I(i)| − |I (i)| . |H| |I(−1)| P|PF i≥0 P|p

Here are some hints for the above equations:

(a): Exactly nPF places of L induce the same summand d(PF |p) deg(PF ).

14 3.1. The general case

(d): cf. Hilbert’s Diﬀerent Formula (e): Both sums are ﬁnite. |G| (f): The number of places of L above p is |I(−1)| , cf. Stichtenoth [47, III.8.2].

P Next we transform the expression P|p |IP|PF (i)|. Let T be a left-transversal for H in G. We have X (A) 1 X |IP|P (i)| = Iπg|(πg) (i) F |I(−1)| F P|p g∈G (B) |H| X = Iπt|(πt) (i) |I(−1)| F t∈T

Here again some hints, that explain the transformations in detail: (A): G acts transitively on the set of places of L lying above p. By definition the decomposition group I(−1) equals the stabilizer of π. Thus, there exist exactly |I(−1)| elements in G that map π to a fixed place P lying over p. (B): Let h ∈ H be an element of H. As h fixes F pointwise, the definitions immediately give th t h t th th t t (π )F = (π )F = (π )F . Since L|F is Galois, the equality Iπ |(π )F (i) = Iπ |(π )F (i) holds. t t t t t t (C): Because of Iπ |p(i) = Iπ|p(i) = I(i) and Iπ |(π )F (i) = Iπ |p(i) ∩ H it follows t −1 t t t Iπ |(π )F (i) = Iπ |p(i) ∩ H = I(i) ∩ H = I(i) ∩ tHt . (D): Let g ∈ I(i). g fixes the coset tH if and only if (tH)g = g−1tH = tH ⇐⇒ t−1g−1t ∈ H ⇐⇒ g−1 ∈ tHt−1 ⇐⇒ g ∈ tHt−1. −1 P −1 Thus, the stabilizer of tH in I(i) coincides with I(i)∩tHt and t∈T I(i)∩tHt equals the sum of the number of fixed points of all elements of I(i). Therefore the orbit-counting theorem yields X −1 I(i) ∩ tHt = |I(i)| · o I(i) . t∈T Putting all together we obtain X deg(π) X |G| X d(q|p) deg(p) = |I(i)| − |I (i)| |H| |I(−1)| P|PF q|p i≥0 P|p deg(π) X |G| |H| = |I(i)| − |I(i)| · oI(i) |H| |I(−1)| |I(−1)| i≥0 deg(π) X = |I(i)| n − oI(i) |I(−1)| i≥0 X n − o I(i) = deg(p) = ind(p) [I : I(i)] i≥0

15 3. Calculation of diﬀerents and the genus-0 condition

The remaining assertions follow from the general deﬁnitions, cf. [47, III.4.3].

The next lemma gives a lower bound for the function “ind”. An obvious idea is to discard all summands with i > 0; this method yields ind(p) ≥ n − o(I). However, the following estimation is stronger.

Lemma 3.5 With the notation from Proposition 3.4 set p the characteristic of E and deﬁne o0(I) the number of orbits of the inertia group I := I(0) with length not divisible by p. Then

ind(p) ≥ n − o0(I).

Proof. Denote by `(t) := |(tH)I | the length of the I-orbit through tH. [47, III.1.6] and the identity Fix(t−1Ht) = F t show |I| e(π|p) `(t) = −1 = = e(πF t−1 |p). |I ∩ tHt | e(π|πF t−1 ) Dedekind’s Diﬀerent Theorem yields

`(t) = d(πF t−1 |p) + δt where δt is an integer with δt ≤ 1 and δt = 1 if and only if p - e(πF t−1 |p). Set ( 0 if p | e(π −1 |p), δ0(t) := F t 1 otherwise.

Then deg(π) X ind(p) = e(P|P )d(P |p) |H| F F P|p

deg(π) · |I| X d(PF |p) = |H| e(PF |p) P|p g (α) deg(π) · |I| X d (π )F |p = |H| · |I(−1)| e(πg) |p g∈G F t (β) deg(π) · |I| X d (π )F |p = |I(−1)| e(πt) |p t∈T F (γ) X d πF t−1 |p = deg(p) e π −1 |p t∈T F t X `(t) − δ0 ≥ deg(p) t `(t) t∈T X 0 ≥ deg(p) n − δt t∈T (δ) = deg(p)n − o0(I)

Again some hints for the above equations:

16 3.1. The general case

Now we state two important theorems that allow us to give severe restrictions for the decrease in the orders of the higher ramification groups. We continue to assume L|E to be Galois and suppose the constant field of E to be algebraically closed. p ∈ P(E) is a place of E and P ∈ P(L) lies over p. For an integer i set I(i) the i-th ramification group of P|p. For a real number u ∈ R define ( I(−1) if u < −1, Iu := I(due) if u ≥ −1. + Let ϕP|p : R0 → R be the function defined by Z u dt ϕP|p(u) := . 0 [I0 : It]

Since its derivative is always positive, ϕP|p is injective. The Theorem of Hasse-Arf [44, IV §3] now reads as follows

Theorem 3.6 (Hasse-Arf) If I(0) is an abelian group and i ∈ N0 is an integer with I(i) > I(i + 1), then ϕP|p(i) ∈ Z. Suppose N is a normal subgroup of Gal(L|E). Set F := Fix(N) and p0 := P ∩ F . Denote 0 0 the ramification groups of P|p with J(i) resp. Ju and the ramification groups of p |p with F (i) resp. Fu. It is easy to express Ju in terms of Iu; the definitions directly give

Ju = Iu ∩ N.

The computation of Fu is more difficult; however, [44, IV §3] shows Theorem 3.7 (Herbrand) With the above notation it follows F =∼ I N/N =∼ I /J . ϕP|p0 (u) u u u Remark 3.8 Although Serre [44] proves the above results only in case of local function fields, they are correct in the global case, too. This is seen as follows: Completion at the place P gives an Galois extension L0|E0 with Galois group I(0). The normal subgroup N of Gal(L|E) corresponds to the group J(0) = I(0) ∩ N. In particular, the function ϕ is invariant under completion as it only depends on subgroups of I(0) resp. J(0). Since, moreover, completion at P does not change the ramification behavior of P|p, the assertions of the Theorems of Hasse-Arf and Herbrand can be transferred unmodified to the case of global function fields. ?

17 3. Calculation of diﬀerents and the genus-0 condition

3.2. Applications to the case of aﬃne Galois groups

In this section p ∈ P is a prime, k an algebraic extension of Fp, t a transcendental over k, and f ∈ k[X] an affine polynomial of degree n = pr that is functionally indecomposable over k. Set ` the splitting field of f(X) − t|k(t). Denote by K an algebraic closure of k and set L := K`. Then L|K(t) is a Galois extension with affine Galois group G = N o H; here N denotes the affine kernel of Gal`|k(t) and H is the stabilizer of a root x of f(X) − t.

As the ﬁxed ﬁeld Fix(H) of H coincides with K(t, x) = K(x) and the action of G on the roots of f(X) − t is equivalent to the natural action of G on the left-coset space G/H, Proposition 3.4 and the Riemann-Hurwitz genus formula imply the important

Corollary 3.9 (Genus-0 Condition) With the notation from Proposition 3.4 the equality P p ind(p) = 2n − 2 holds. For a place p ∈ PK(t) of K(t) the integer ind(p) only depends on the series of higher ramiﬁcation groups of p and the way how these groups act on the set of zeros of f(X) − t. In the following sections we present methods that give estimations for these information.

3.2.1. Consequences of Hasse-Arf and Herbrand Set E := Fix(N), fix a place P ∈ P(L), and define p0 := P ∩ E. 0 0 Let Iu resp. Ju resp. Fu denote the ramification groups of P|p resp. P|p resp. p |p. Define I := I0, J := J0, and F := F0. For an integer i ∈ N0 set

−1 ui := ϕP|p0 (i).

Important facts concerning the elements ui are given by the following

Lemma 3.10 ui is an integer. For a real number x ∈ R with ui < x ≤ ui+1 the equalities

Jx = J(ui+1) and Ix = I(ui+1) hold. Furthermore |J| ui+1 − ui = . |J(ui+1)|

Proof. Let m ∈ N0 be an integer and u ∈ R a real number with m < u ≤ m + 1. Since ϕP|p0 is continuous and piecewise linear by deﬁnition, it follows

m |J(m + 1)| X |J(k)| ϕ 0 (u) = (u − m) + . P|p |J| |J| k=1

Thus, assuming m < ui ≤ m + 1 yields

m |J(m + 1)| X |J(k)| ϕ 0 (u ) = i = (u − m) + P|p i |J| i |J| k=1 which is equivalent to m X i|J| = |J(m + 1)|(ui − m) + |J(k)|. k=1

18 3.2. Applications to the case of aﬃne Galois groups

Since |J(m + 1)| divides |J(k)| for all k ∈ {0, 1, . . . , m + 1}, ui − m is an integer and, hence, ui ∈ N0. The ramiﬁcation groups of P|p0 are subgroups of J ≤ N and, thus, abelian. As i < ϕP|p0 (x) ≤ i + 1 and ϕP|p0 (x) = i + 1 if and only if x = ui+1, Hasse-Arf gives

Jx = J(ui+1).

The deﬁnition of u gives F = F (dϕ 0 (x)e) = F (i + 1). Herbrand states i ϕP|p0 (x) P|p

I /J = I /J(u ) =∼ F = F (i + 1); x x x i+1 ϕP|p0 (x) this shows |Ix| = |I(ui+1)| for all possible x. Hence, Ix = I(ui+1). 0 |J(ui+1)| As ϕP|p0 (x) = |J| , the deﬁnitions of ui and x give

|J(ui+1)| ϕ 0 (x) = i + (x − u ). P|p |J| i The remaining assertions follow easily.

Corollary 3.11 The equality

∞ X |F (i)| ind(p) = n − oI(u ) |F | i i=0 holds. |J| Proof. By Lemma 3.10 the ui+1 − ui = groups |J(ui+1)|

I(ui + 1),I(ui + 2),...,I(ui+1) are equal. J being a p-group gives J = J(0) = J(1), u0 = 0, and u1 = 1. Therefore ∞ X n − o I(i) ind(p) = [I : I(i)] i=0 ∞ n − I(u0) X n − o I(ui) = + (u − u ) [I : I(u )] [I : I(u )] i i−1 0 i=1 i ∞ n − I(u0) X |I(ui)| · |J| = + n − oI(u ) . [I : I(u )] i |I| · |J(u )| 0 i=1 i ∼ ∼ Herbrand shows that I/J = F and I(ui)/J(ui) = F (i). Thus, ∞ n − I(u0) X |F (i)| ind(p) = + n − oI(u ) [I : I(u )] i |F | 0 i=1 ∞ X |F (i)| = n − oI(u ) . |F | i i=0 This is the claim.

19 3. Calculation of diﬀerents and the genus-0 condition

3.2.2. Bounds for the number of fixed points for elements of G We have seen in the previous paragraphs that we need information about the number of orbits oI(i) of a series of ramification groups I(i) in order to calculate degrees of differents in the extension K(x)|K(t). As the orbit-counting theorem relates the number of orbits of a group with the number of fixed points of each group element, we have to develop good bounds for this latter number. For an arbitrarily given permutation group this is a difficult problem. In our case G is an affine group; we have geometric interpretations of the action of G. This allows us to give severe restrictions for the number of fixed points of an element of G.

Every element g ∈ G can uniquely expressed in the form g = nh with n ∈ N and h ∈ H. Deﬁne the H-projection mapping L : G → H (the letter “L” stands for “linearization”) via

(nh)L := h.

L is an epimorphism with kernel N. We state an important observation.

Lemma 3.12 Let g ∈ G be an element of G. Then either g fixes no element or g and gL have the same number of fixed points. Proof. Suppose f ∈ N is fixed by g. As N is a transitive subgroup of G, every element fixed by g can be written in the form fm with m ∈ N. Thus, let m ∈ N be an arbitrary element of N and write g = nh with n ∈ N and h ∈ H. Then fm is fixed by nh if and only if fm = (fm)nh = (fm · n)h = (fn · m)h = f nh · mh = f · mh ⇐⇒ m = mh. | {z } ∈N It follows that g has as many fixed points as gL = h.

Since an element h ∈ H acts as an automorphism Ah ∈ GL(N) of the vector space N, the set of ﬁxed points of h equals the eigenspace of Ah for the eigenvalue 1. As eigenspaces are vector spaces, we get

Corollary 3.13 Suppose g ∈ G has at least one fixed point. Then the number of fixed points n of g is a power of p; in particular, 1 6= g implies | Fix(g)| | p . Next we state a criterion that guarantees the existence of fixed points.

Corollary 3.14 Let g ∈ G be a p-regular element. Then g has a ﬁxed point.

Proof. The group U := hN, gi can be written as U = N o C with a cyclic group C ≤ H of order |g|. As N is solvable, Schur-Zassenhaus shows that g is conjugate to a generator of C. This element, however, ﬁxes 0 ∈ N.

3.2.3. Bounds for ind(∞) Let ∞ ∈ PK(t) be the infinite place of K(t) and denote with P ∈ P(L) a place of L lying over ∞. Set I∞(i) the i-th ramification group of P|∞ and define I∞ := I∞(0). The next lemma is the basis for all estimations of ind(∞).

20 3.2. Applications to the case of aﬃne Galois groups

Lemma 3.15 Both I∞ and I∞(1) are transitive. Proof. As ∞ ramiﬁes totally in the extension K(x)|K(t), van der Waerden [51] gives the transitivity of I∞. I∞(1) equals the normal p-Sylow subgroup of I∞. Denote by Ω1,..., Ωr the orbits of I∞(1). Pr As I∞(1) is normal in I∞, all orbits Ωi have the same cardinality ω. Since |N| = i=1 |Ωi| is a power of p, both ω and r are powers of p. I∞ acts transitively on the set of orbits Ωi. Hence, r divides [I∞ : I∞(1)]; as this index is prime to p, we obtain r = 1 and, thus, the transitivity of I∞(1).

Easy consequences are

r −1 Corollary 3.16 Suppose |I∞| = p s with p - s. Then ind(∞) ≥ (n − 1)(1 + s ); equality holds if and only if I∞(2) = 1. In particular, ind(∞) ≥ n and, provided that I∞ > 1 is a p-group, ∞ is the unique place ramifying in L|K(t).

n Corollary 3.17 If I∞(i) > 1, then o I∞(i) | p . Use the notation from Corollary 3.11 and suppose I∞(ua) > I∞(ua+1) = 1. Then

a |F∞(1)| n X |F∞(i)| ind(∞) ≥ (n − 1) 1 + + n − . |F | p |F | ∞ i=2 ∞

Proof. Every ramiﬁcation group I∞(i) is a normal subgroup of I∞, cf. Maus [37]. Thus, the orbits of I∞(i) all have the same length. The claim is due to n being a prime power.

The previous results and our deﬁnition E := Fix(N) give

Lemma 3.18 ∞ ramiﬁes in E|K(t) if and only if [E : K(t)] > 1.

Proof. Suppose ∞ does not ramify in the extension E|K(t). Then I∞ ≤ N. Since ∞ ramifies totally in the extension K(x)|K(t), the group I∞ coincides with N. By Corollary 3.16 there is no finite ramification in L|K(t). Therefore E|K(t) is unramified. Stichtenoth [47, III.5.8] shows E = K(t).

3.2.4. Bounds for ind(p) with p a finite place 0 Let p ∈ P K(t) be a finite place of K(t) and P ∈ P(L) lying over p. The symbols p , Ip, Jp, and Fp are defined as in section 3.2.1. Lemma 3.5 allows us to prove an important

∼ L Corollary 3.19 Jp = 1. Ip(i) = Fp(i) and o Ip(i) ≤ o Ip(i) for all i.

Proof. Suppose Jp is nontrivial. Then p | |Jp| and every Jp-orbit has length a multiple of p. As Jp is a subgroup of Ip, every Ip-orbit is the disjoint unit of some of the Jp-orbits. In particular, every Ip-orbit has length divisible by p. Lemma 3.5 shows

ind(p) ≥ n.

This contradicts the genus-0 condition as also ind(∞) ≥ n.

21 3. Calculation of diﬀerents and the genus-0 condition

Thus, Jp = 1 and ϕP|p0 is the identity. Herbrand gives ∼ Fp(i) = Ip(i)/Jp(i) = Ip(i).

The remaining assertion follows from the orbit-counting theorem.

We see that the knowledge of the ramiﬁcation of p in the extension E|K(t) determines the ramiﬁcation of p in L|K(t) completely. Furthermore the estimation

L X n − o Ip(i) ind(p) ≥ [Ip : Ip(i)] i≥0 holds.

A consequence of the orbit-counting theorem is

Lemma 3.20 Set f := max{| Fix(g)| | g ∈ Ip(i)} and suppose 1 ≤ r < |Ip(i)|. Then

1 1 n 1 n o Ip(i) ≤ n + (|Ip(i)| − 1)f ≤ n + (|Ip(i)| − 1) ≤ n + (r − 1) . |Ip(i)| |Ip(i)| p r p A simple but useful consequence of the previous estimations is the following lemma that gives a limit for the number of finite branch points for affine polynomials. This result is also proved in Guralnick/Müller [17, Lem. 2.1].

Lemma 3.21 Suppose f is an affine polynomial of degree pr. If f has ≥ 2 finite branch points, then f has exactly 2 finite branch points, p is odd, and the finite branch points are tamely ramified with corresponding inertia groups of order 2.

22 Part II.

Application to speciﬁc problems

Notation used in this part

We continue the notation from the previous sections: k is a finite field of characteristic p. f ∈ k[X] denotes an affine polynomial of degree n = pr that is functionally indecomposable over k. K is a fixed algebraic closure of k. t is a transcendental over K, x denotes a fixed zero of f(X) − t. The splitting field of f(X) − t over k(t) resp. K(t) is denoted by ` resp. L = K`. The arithmetic monodromy group Gal`|k(t) of f is denoted by A, the geometric monodromy group Gal L|K(t) by G. N is the affine kernel of A. Set U := Ax and H := Gx. The fixed field of N in the extension L|K(t) is denoted by E. ∞ always stands for the infinite place of K(t); the symbols p and q denote different finite places of K(t). Unless redefined, I∞(i) is the i-th ramification group of a fixed place of L over ∞, J∞(i) := I∞(i) ∩ N, and F∞(i) denotes the i-th ramification group of the induced extension of ∞ in E|K(t). Set I∞ := I∞(0), J∞ := J∞(0), and F∞ := F∞(0). The groups Ip, Iq, etc. are defined analogously.

4. Aﬃne polynomials with g ≤ 2

In this chapter we restrict the genus g := g(E) of E to certain values and classify all aﬃne polynomials in question. A ﬁrst observation is

Lemma 4.1 Suppose the extension E|K(t) is tame. Then g = 0 and the possibilities for f are given in Theorem 4.2. Proof. The assertion is clear for E = K(t).

Otherwise ∞ ramiﬁes in E|K(t) with index e∞ := |F∞| > 1 by Lemma 3.18. Additionally at most two ﬁnite places p and q ramify; set ep := |Fp| and eq := |Fq|. Suppose eq = 1. Then by Riemann-Hurwitz e − 1 e − 1 2g − 2 = −2[E : K(t)] + ∞ · [E : K(t)] + p · [E : K(t)] < 0; e∞ ep | {z } <2[E:K(t)] hence g = 0. Suppose ep, eq > 1. Then ep = eq = 2 by Lemma 3.21 and Riemann-Hurwitz shows e − 1 1 2g − 2 = −2[E : K(t)] + ∞ · [E : K(t)] + 2 · · [E : K(t)] < 0. e∞ 2 Thus, the claim follows in in this case, too.

4.1. g = 0

In this section we assume the ﬁeld E to be of genus 0. Our main result will be

Theorem 4.2 With the notation from page 23 suppose g = 0. Then there exist linear polynomials g1, g2 ∈ K[X] such that F := g1 ◦ f ◦ g2 ∈ k[X] belongs to one of the following classes:

Pr pi (1) H = 1 and F is an additive polynomial, i.e. F = i=0 aiX . F is exceptional if and only if F has no nonzero root in k.

(2) Case (A) of Theorem 4.3 holds and F is sublinearized of the form (4.2), cf. page 28. q m 1 F is exceptional if and only if the polynomial X F (X) has no zero in k. (3) Case (D) of Theorem 4.3 holds. In [18] F will be studied in detail.

(4) Case (E) of Theorem 4.3 holds. F fulﬁlls the conclusion of [17, Theorem 1.4].

25 4. Aﬃne polynomials with g ≤ 2

We start by describing the Galois theoretic structure of the extension E|K(t). This is not diﬃcult since the possible Galois groups and the corresponding ramiﬁcation data of E|K(t) are well known.

Theorem 4.3 E is a rational function field. The following table gives all possible isomor- phisms types for H with E|K(t) having the associated ramification behavior. Case H =∼ Ramification data Conditions

(A) Cm (m, m) (p, m) = 1 (B) Cp × · · · × Cp (|H|) (C) (Cp × · · · × Cp) o Cm (m, qm) |H| = qm, m | q − 1, q 6= 1 (D) Dm (2, m) p = 2, (2, m) = 1 (E) Dm (2, 2, m) p 6= 2, (p, m) = 1 (F) A5 (5, 6) p = 3 q(q−1) q+1 m (G) PSL(2, q) 2 , 2 p 6= 2, q = p (H) PGL(2, q) q(q − 1), q + 1 p 6= 2, q = pm (I) PGL(2, q) q(q − 1), q + 1 p = 2, q = pm Proof. As K is algebraically closed, the rationality of E is a direct consequence of Stichtenoth [47, I.6.3]. Thus, E|K(t) is a Galois extension of rational function fields. The ramification behavior and the Galois groups of such extensions were classified by Valentini and Madan in [50]. Some cases of the original classification cannot occur: on the one hand, K(t) does not contain any place of degree > 1, on the other hand, an affine polynomial fulfills the conclusion of Lemma 3.21. This leads to the above cases.

The case H = 1 is completely discussed by Proposition 2.9. The criterion for an additive polynomial to be exceptional is given for instance in Lidl/Niederreiter [35, Theorem 7.9]. Therefore we will assume H > 1 from now on.

∼ 4.1.1. Case (A): H = Cm is cyclic

As ∞ ramifies totally in the extension K(x)|K(t), mn is a divisor of |I∞|. Since G has order mn, the equality I∞ = G holds. We show that no nontrivial element of H fixes an element of N ]. Assume the contrary. Then there exist an element a ∈ N ] and a nontrivial subgroup S ≤ H such that a is fixed by every element of S. As H is cyclic, S is characteristic in H and, thus, normal in U. Set M := hai the irreducible S-submodule of N generated by a. Then every element of M is fixed by S. Clifford allows us to decompose the S-module N in the form

s M N = M ui i=1 where ui ∈ U are appropriate elements of U. It follows that S stabilizes every element of M ui . Hence, S is a subgroup of the kernel of the operation of H on N in contradiction to the faithfulness of this operation. Corollary 3.14 shows 1 1 ind(p) = n − n + (m − 1) · 1 = (n − 1) 1 − . m m

26 4.1. g = 0

The above shows that a conjugate of Ip either is a subgroup of H or intersects H trivially. Thus, places of K(x) lying over p either ramify with index m or index 1. Suppose there are a resp. b places of K(x) lying over p with index m resp. index 1. By solving the equations am + b = n and a(m − 1) = ind(p) we obtain that p decomposes in K(x) in the form m m p = P · P1 ··· P n−1 with pairwise diﬀerent places P, Pi ∈ P K(x) . (4.1) m The genus-0 condition enforces 1 n − o I∞(1) ind(∞) = 2n − 2 − ind(p) = (n − 1) 1 + = n − o(I∞) + ; m [I∞ : I∞(1)] this gives I∞(2) = 1.

Now we are able to compute the genus g(L) of L. Our previous considerations prove that the ramification in the extension L|K(x) comes from the infinite place of K(x) and P; both places ramify totally. Hence, by Riemann-Hurwitz g(L) = 0 and L is a rational function field. Construction of f By a linear substitution of both t and x we may assume p resp. P to be the zero place of K(t) resp. K(x). The decomposition (4.1) of p gives f(X) = X · gm(X) n−1 with a separable polynomial g ∈ K[X] of degree m and g(0) 6= 0. The derivative of f reads as f 0(X) = gm(X) + mXgm−1(X) · g0(X) = gm−1(X) · g(X) + mX · g0(X). Since p is the only finite place ramifying in K(x)|K(t), every zero ξ of f 0 is also a zero of g. Assume g(ξ) + mξg0(ξ) = 0. Then mξg0(ξ) = 0. Since mξ 6= 0, we get g0(ξ) = 0. But this is impossible as g does not have multiple roots. Hence, g(X) + mXg0(X) ∈ K] is a nonzero element of K. n−1 P m i Write g(X) = giX with gi ∈ K and g0 · g n−1 6= 0. We obtain the equivalence i=0 m

n−1 m 0 X i ] g(X) + mX · g (X) = gi(1 + mi)X ∈ K ⇐⇒ gi(1 + mi) = 0 for i ≥ 1. i=0

This shows that for i ≥ 1 a nonzero coefficient gi 6= 0 is possible only if 1 + mi = 0, i.e. 1 + mi = αp with some α ∈ N. Hence, X αp−1 m f(X) = X g0 + g αp−1 X m . m m|αp−1 αp−1 n−1 1≤ m ≤ m As L is a rational function field, it is possible to choose z ∈ L such that L = K(z) and the infinite resp. zero place of K(z) lies over the infinite resp. zero place of K(x). Hence, we may assume zm = x. Then z is annihilated by

:=F (X) z }| { m m X αp m f ◦ X − t = g0X + g αp−1 X − t = X ◦ F (X) − t. m m|αp−1 αp−1 n−1 1≤ m ≤ m

27 4. Aﬃne polynomials with g ≤ 2

Set τ := F (z). As τ m = t, the field K(τ) contains K(t). Since deg F = n, the extension K(z)|K(τ) has degree [K(z): K(τ)] = n. N is the unique subgroup of G of order n; thus, the equality Fix(N) = K(τ) holds. As the infinite place of K(τ) ramifies totally in the extension K(z)|K(τ) with the infinite place of K(z) lying above, Proposition 2.9 shows that F is an additive polynomial. Thus, the indices α are powers of p. We obtain

X αp−1 m X pi−1 m m m f(X) = X g0 + g αp−1 X = X g pi−1 X . (4.2) m m m|αp−1 m|pi−1 αp−1 n−1 i 1≤ m ≤ m p |n This are exactly Cohen’s sublinearized polynomials, cf. [9]. We state Cohen’s criteria for f to be exceptional in Theorem 4.2.

4.1.2. Case (B): H is an elementary abelian p-group H is a normal p-subgroup of U. Proposition 2.8 gives H = 1 which was previously excluded.

∼ 4.1.3. Case (C): H = (Cp × · · · × Cp) o Cm is a semidirect product H contains a unique p-Sylow subgroup. Thus, the p-Sylow subgroup of H is normal in U. Hence, H is a p0-group in contradiction to Theorem 4.3

4.1.4. Case (D): H is dihedral in even characteristic This case will be extensively studied in [18]. It will be proved that there exist exceptional polynomials realizing this case.

4.1.5. Case (E): H is dihedral in odd characteristic As G/N is a p0-group and f has two finite branch points, Guralnick/Müller[17, Theorem 2.2] completely discusses the ramification of L|K(t). It will be shown in [18] that this case leads to the class of polynomials described in [17, Theorem 1.4].

∼ 4.1.6. Case (F): H = A5 in characteristic 3 We show that this case does not occur.

3 Suppose ﬁrst that p ramiﬁes tamely. Then by Corollary 3.16 ind(∞) ≥ 2 (n − 1) and 1 n ind(p) ≥ n − n + 4 · . 5 3 This violates the genus-0 condition.

∼ ∼ Now suppose that p ramiﬁes wildly. Then I∞ = N o C5 and Ip = S3; the higher ramiﬁ- cation groups of ∞ are subgroups of N and, thus, act semiregularly on N. It follows from section 3.2.1 ∞ ∞ X |F (i)| 6 n X 1 ind(∞) = n − oI (u ) = (n − 1) + 1 − . |F (0)| ∞ i 5 5 |I (u )| i=0 i=2 ∞ i

28 4.1. g = 0

Therefore 6 4 6 ind(∞) = (n − 1) or ind(∞) ≥ n − 5 3 5 depending on whether I∞(2) = 1 or I∞(2) ≥ C3 > 1. Since elements of the same order are conjugate in Ip, ind(p) is given by

1 1 n − (n + 2f3) ind(p) = n − (n + 3f + 2f ) + s · 3 6 2 3 2 where f2 resp. f3 denotes the number of fixed points of an element of Ip of order 2 resp. 3 n and s is the well-defined integer with Ip(s) > Ip(s + 1) = 1. As both f2 and f3 are ≤ 3 , the 6 n genus-0 condition immediately gives ind(∞) = 5 (n − 1), s = 1, and f2 = f3 = 3 . It follows 36 − n ind(∞) + ind(p) − 2n + 2 = ; 45 hence, the genus-0 condition cannot be fulfilled.

4.1.7. Case (G): H ∼= PSL(2, q) with p 6= 2 We prove that this case does not occur. H is irreducible on N We show ﬁrst that we can construct a new aﬃne polynomial g ∈ K[X] from our given polynomial f such that g is indecomposable over K and the geometric point stabilizer of g is isomorphic to PSL(2, q).

Lemma 4.4 Suppose H is a non-abelian group such that every normal subgroup of H is characteristic in H. Then there exists an affine polynomial g ∈ K[X] that is indecomposable over K and whose geometric point stabilizer is isomorphic to a factor group of H. Moreover, the fixed field of the affine kernel of the geometric monodromy group of g is a rational function field. Proof. If f is functionally indecomposable over K, then we can set f = g. Otherwise H is reducible on N and we can find a proper H-submodule M < N. Since H is normal in U, N is a semisimple H-module and can be written in the form s M ui ] N = M with u1 = 1 and ui ∈ U . i=1 H is faithful on M because the kernel of the action of H on M is characteristic in H and, thus, a subgroup of the kernel of the action of H on N. As H is non-abelian, this shows in particular that M cannot be one-dimensional. 0 L ui 0 0 Set N := i6=1 M ; N is an H-module with H < N o H < N o H. By Galois duality 0 Fix(N oH) = K(y) is a proper intermediate field between K(x) and K(t). This field induces a decomposition f = g ◦ h over K with h(x) = y and g(y) = t, cf. Lemma 2.2. T 0 σ Set Z the Galois hull of K(y)|K(t) and J := Gal(L|Z); by definition J = σ∈G(N o H) . 0 0 0 0 N is invariant under the action of H; hence, N ≤ J ≤ N o H. This shows that J = N o Q with Q a normal and, thus, characteristic subgroup of H. The extension Z|K(t) is Galois by construction; the Galois group of Z|K(t) is given by

∼ N o H ∼ 0 ∼ Gal Z|K(t) = G/J = = N/N o H/Q = M o H/Q. N 0 o Q

29 4. Aﬃne polynomials with g ≤ 2

N H ∼ 0 0 The isomorphism 0o = N/N H/Q comes from the fact that every element nhN Q ∈ N oQ o N H 0 0 0 0o can be written in the form nN · hQ since hN = N h. N oQ Since there are no proper intermediate ﬁelds between K(y) and K(t), g is indecomposable over K and H/Q acts irreducibly on M.

GalZ|K(t) is an affine group: the regularity of N on the set of zeros of f(X) − t gives the regularity of N/N 0 on the set of zeros of g(X) − t. The fixed field of the geometric point stabilizer of g equals the fixed field of N o Q and, thus, is a subfield of the rational function field E. Lüroth [47, III.5.9] shows that this field is rational, too.

The following diagram visualizes the above situation

L 1 oo OOO oo OOO ooo OOO ooo OOO oo OO ooo OOO o OO oo 0 O K(x) Z E H N o Q N / ?? / ? / ? // ?? / ? / ? // / / Fix(N o Q) / N o Q / // // / / 0 K(y) N o H ? ? ?? ?? ? ? K(t) N o H

For q > 3 the group PSL(2, q) is simple. Thus, the geometric monodromy group of g is MoH/Q with H/Q ∈ {1, PSL(2, q)}. The case H/Q = 1 is impossible because the irreducible H/Q-module M would be one-dimensional in contradiction to Lemma 4.4. ∼ ∼ Suppose q = p = 3. Then H = PSL(2, 3) = A4 and every normal subgroup of H is also characteristic in H. We obtain H/Q ∈ {1,C3,H}; this again yields H/Q = H.

Hence, in all cases we can construct a functionally indecomposable polynomial g from f that also belongs to case (G). Therefore we will assume from now on that the given polynomial f itself is functionally indecomposable. By showing that such an f does not exist we prove the impossibility of case (G). Bounds for the number of ﬁxed points for elements of H The following concept allows us to obtain approximate upper bounds for the number of ﬁxed points of elements of H.

Deﬁnition 4.5 Let H be a group and h ∈ H. We say that h is k-generating in H if H can be generated by k conjugates of h, i.e. if there exist elements g1, . . . , gk ∈ H such that

H = hhg1 , . . . , hgk i.

This deﬁnition yields in case of the group PSL(2, q)

Lemma 4.6 Let 1 6= x ∈ PSL(2, q) with q any prime power (in particular, the case q = 2m is allowed). Then x is • 2-generating if q > 2, |x| > 2, and (|x|, q) 6= (3, 9),

30 4.1. g = 0

• 3-generating if (|x|, q) = (3, 9),

• 3-generating if |x| = 2 and q 6= 3.

Proof. The case (|x|, q) = (3, 9) follows directly from Guralnick/Saxl [21, Lem. 3.1]. Now suppose |x| = 2 and q 6= 3. [21] gives the assertion for q ≥ 4. For q = 2 the claim follows by direct computation.

At last suppose q > 2, |x| > 2, and (|x|, q) 6= (3, 9). If |x| is not a power of 2, then the assertion follows for q ≥ 4 again from [21] and for q = 3 by direct computation. Thus, assume |x| = 2i with i ≥ 2. It is suﬃcient to show that an element f ∈ PSL(2, q) of order 4 is 2-generating. An inspection of the list [25, II 8.27] of all subgroups of PSL(2, q) shows that for q 6∈ {7, 9} the group PSL(2, q) contains a maximal subgroup D isomorphic to a dihedral group q+1 of order 2 · (2,q−1) . The well-known fact that PSL(2, q) contains exactly one conjugacy class of involutions allows to ﬁnd elements g, h ∈ PSL(2, q) such that D = h(f 2)g, (f 2)hi. Suppose hf g, f hi= 6 PSL(2, q). Then hf g, f hi ≤ D, and both f g and f h are elements of the normal 1 g 2 h 2 cyclic subgroup of D of order 2 |D|. This implies (f ) = (f ) in contradiction to D being dihedral. For q ∈ {7, 9} the assertion follows by direct calculation.

The next lemma gives an upper bound for the number of ﬁxed points of a k-generating element.

Lemma 4.7 Let h ∈ H be an element of H. If h is k-generating in H, then

k − 1 √k dim Fix(h) ≤ dim N and | Fix(h)| ≤ nk−1. k

Proof. Let h ∈ H be k-generating in H. Then there exist conjugates h1, . . . , hk of h that generate H. Set Fi := Fix(hi), f := dim F1, and d := dim N; obviously the equality f = dim Fi holds for all i ∈ {1, . . . , k}. Consider the homomorphism

ϕ : N → N/F1 ⊕ · · · ⊕ N/Fk, n 7→ (nF1, . . . , nFk)

Tk Tk with kernel ker ϕ = i=1 Fi. As H is irreducible on N, the intersection i=1 Fi = {0} is trivial and ϕ is injective. Since dim N/Fi = d − f, we obtain

k − 1 d ≤ k · (d − f) ⇐⇒ f ≤ d. k The remaining assertions follow easily.

N is a “big” module: n ≥ q4 For this section we assume n ≥ q4.

31 4. Aﬃne polynomials with g ≤ 2

Suppose q 6= 9 and h ∈ H. Lemmas 4.6 and 4.7 give the estimations  n for h = 1, √  3 2 n n | Fix(h)| ≤ n = √3 ≤ for |h| = 2 and q 6= 3, √ n q  √n n  n = n ≤ q2 for |h| > 2.

q+1 First we discuss the case where p ramiﬁes tamely with index . Then I∞ = P o C q−1 with 2 2 a transitive p-group P . Corollary 3.16 shows 2 ind(∞) ≥ (n − 1)1 + ; q − 1 this contradicts the genus-0 condition if q = 3. Ip is cyclic and, thus, contains at most one involution. Hence, 2 n q + 1 n ind(p) ≥ n − n + + − 2 . q + 1 q 2 q2 It follows q2(q + 1)(q − 3) + n−3 + q(6 + q) ind(∞) + ind(p) − 2n + 2 ≥ > 0; q4 − q2 so the genus-0 condition is violated for all q in question.

q(q−1) ∼ Next we consider the case where p ramiﬁes wildly with index . Here, I∞ = N o C q+1 2 2 2 and, thus, ind(∞) ≥ (n − 1) 1 + q+1 . The following lemma allows us to estimate the number of involutions in Ip. ∼ Lemma 4.8 Suppose G = AoB with 2 - |A| and B cyclic. Let i be the number of involutions and f the number of elements of order 4 in B. Then the number of involutions resp. elements of order 4 in G is less or equal than i · |A| resp. f · |A|. Proof. Let a ∈ A and b ∈ B. As A is a normal 20-subgroup of G, it follows by some standard arguments that ab ∈ G having order 2 resp. 4 implies b to be an involution resp. to be of order 4. The claim follows as a direct consequence of this fact.

As Ip is isomorphic to a product P o C q−1 with a p-group P of order q, the above lemma 2 0 shows that Ip contains at most q involutions. Since Ip is a 2 -group for q = 3, we obtain in all cases 2 n q(q − 1) n 2 1 n ind(p) ≥ n − n + q + − 1 − q + n − n + (q − 1) . q(q − 1) q 2 q2 q − 1 q q2 These estimations succeed in disproving the genus-0 condition.

Now suppose q = 9. If p ramiﬁes tamely, the same bounds as above hold and, hence, the same contradiction follows. Otherwise Ip contains exactly 9 involutions and 8 elements of order 3; this shows 1 n n 1 n o(I ) ≤ n + 17 + 18 and oI (1) ≤ n + 8 . p 36 9 81 p 9 9 Again a contradiction to the genus-0 condition results.

32 4.1. g = 0

N is a “small” module: n < q4 Unfortunately, in this case the bounds from Lemma 4.7 are too weak to induce contradictions to the genus-0 condition. By classifying all irreducible representations of H on N we get much sharper estimations for the number of ﬁxed points of elements of H, cf. Lemma 3.12. This method will eventually lead to the impossibility of this case.

The following presentation of p-modular representation theory of the groups PSL(2, q) and PGL(2, q) is based on Brauer and Nesbitt [5, §30].

Let κ be any field of characteristic p and u, v algebraically independent transcendentals over κ. Denote by Vn the vector space of homogeneous polynomials in u and v of degree n over n n−1 n κ. Obviously Bn := (u , u v, . . . , v ) is a basis for Vn. a b For an element A = c d ∈ GL(2, κ) define uA := au + bv and vA := cu + dv. Some calculation shows that for a second element A0 ∈ GL(2, κ) the relations (uA)A0 = uAA0 and (vA)A0 = vAA0 hold. This action of GL(2, κ) can be extended to an endomorphism ϕA of Vn by defining n n ϕA i n−i ϕA A i A n−i X i n−i X A i A n−i (u v ) := (u ) (v ) and, thus, κiu v = κi(u ) (v ) . i=0 i=0

The mapping A 7→ ϕA is a homomorphism GL(2, κ) → Aut(Vn). Set Sn(A) the representation matrix of ϕA with respect to the basis B; we consider Sn(A) always acting on the row vector space κn+1. Then

GL(2, κ) → GL(n + 1, κ),A 7→ Sn(A) is a group homomorphism. The next lemma allows us to explicitly calculate Sn(A).

a b Lemma 4.9 Let A = c d ∈ GL(2, κ). Denote the entries of Sn(A) by sij with 1 ≤ i, j ≤ n + 1. Then

min(n+1−i,n+1−j) X n + 1 − i i − 1 s = aλbn+1−i−λcn+1−j−λdλ+i+j−n−2. ij λ n + 1 − j − λ λ=max(0,n+2−i−j) Proof. For the sake of shortness set n0 := n + 1. The image of the i-th basis vector n+1−i i−1 u v of B under the mapping ϕA is 0 0 (un −ivi−1)ϕA = (au + bv)n −i(cu + dv)i−1 0 n −i i−1 0 X X n − i i − 1 0 = arbn −i−rcsdi−1−sur+svn−r−s r s r=0 s=0 0 n min(µ,n −i) 0 (?) X X n − i i − 1 0 = aλbn −i−λcµ−λdi−1+λ−µ uµvn−µ λ µ − λ µ=0 λ=max(0,µ+1−i) In (?) we set µ = r+s and λ = r. The conditions for λ come from the fact that 0 ≤ r ≤ n+1−i and 0 ≤ s ≤ i − 1. This shows the assertion.

33 4. Aﬃne polynomials with g ≤ 2

r×s t×u Deﬁnition 4.10 (Kronecker Product) Let A = (aij) ∈ κ , B ∈ κ . The Kronecker product A ⊗ B of A with B is the matrix   a11B ··· a1sB rt×su A ⊗ B := ......  ∈ κ . ar1B ··· arsB

Now we are able to specify all irreducible p-modular representations of PSL(2, q) and PGL(2, q) over Fq:

m Theorem 4.11 (Brauer-Nesbitt) Let p ∈ P be a prime and q := p a power of p. Let ϕi denote the i-th power of the Frobenius-automorphism of Fq, i.e.

pi ϕi : Fq → Fq, x 7→ x .

ϕ ϕ For a matrix A = (aij) with aij ∈ Fq set A i := (aij) i . (1) Assume p ≥ 3. For A ∈ SL(2, q) denote by A the image of A in PSL(2, q). Then, up to conjugation, the irreducible p-modular representations of PSL(2, q) over Fq correspond to mappings θ with

m−1 m−1 Y O ϕi θ : PSL(2, q) → GL (ri + 1), q , A 7→ Sri (A ); i=0 i=0

Pm−1 here 0 ≤ ri < p for all i and 2 | i=0 ri. θ is faithful if and only if there exists at least one index i with ri 6= 0. (2) For A ∈ GL(2, q) denote by A the image of A in PGL(2, q). Then, up to conjugation, the irreducible p-modular representations of PGL(2, q) over Fq correspond to mappings θ with

m−1 m−1 Y s O ϕi θ : PGL(2, q) → GL (ri + 1), q , A 7→ det(A) Sri (A ); i=0 i=0

Pm−1 i here 0 ≤ ri < p for all i, 0 ≤ s < q − 1, and q − 1 | (2s + i=0 rip ).

Fq is a splitting ﬁeld for both PSL(2, q) and PGL(2, q). We proceed with our classiﬁcation of irreducible H-modules N with n < q4:

We interpret N as a Fp-vector space; then H acts as a subgroup of GL(r, p). Isaacs [29, 9.21] q and Theorem 4.11 show that N F := N ⊗ Fq is completely reducible, i.e.

q N F = V1 ⊕ · · · ⊕ Vf with absolutely irreducible and pairwise non-similar H-modules Vi over Fq. We further obtain from [11, 70.15] that the Vi form an orbit under the operation of the Galois group

p X := Gal(Fq|Fp) = hξ : Fq → Fq, z 7→ z i.

X |X| f Set X1 the stabilizer of V1. Since f = |V | = , we see that X1 = hξ i. 1 |X1|

34 4.1. g = 0

Theorem 4.11 shows that V1 is similar to a representation uniquely parametrized by an m-tuple of integers ri with 0 ≤ ri < p; by abuse of notation set

V1 = (r0, . . . , rm−1).

It follows from the deﬁnitions that

ξ V1 = (rm−1, r0, . . . , rm−2);

thus, the operation of ξ on V1 induces a cyclic right shift in the parametrization of V1. If f we consider the indices of the parameters ri modulo m, the property of ξ stabilizing V1 translates into ri = ri+f for all i; setting g := gcd(f, m) this condition is equivalent to

ri = rj if i ≡ j mod g.

So there exist elements R0,...,Rg−1 with 0 ≤ Ri < p such that

m−1 g−1 g−1 m m g Y Y g Y dimFq V1 = (ri + 1) = (Ri + 1) = (Ri + 1) i=0 i=0 i=0 | {z } =:J and m−1 g−1 X m X r = R . i g i i=0 i=0 m Fq g 4 Since dimFp N = dimFq N = fJ , the condition n < q gives

m g m f m m n = pfJ < p4m ⇐⇒ fJ g < 4m ⇐⇒ J g < 4 . g g

f m b Set a := g and b := g ; then we have to solve aJ < 4b. In particular J > 1 for otherwise V1 would be trivial and H would not be faithful on N. Since a ≥ 1, we get b < 4. Therefore only the following cases need to be discussed: b = 3: We have to solve aJ 3 < 12; this is only possible for (a, J) = (1, 2). But then X X 2 - ri = b Ri = b · 1 = 3.

This shows that the case does not occur. b = 2: We have to solve aJ 2 < 8; this is only possible for (a, J) = (1, 2). It follows f = g and

fJb gJb 4· m 2m 2 n = p = p = p 2 = p = q .

m Since J = 2, there exists an index i, 0 ≤ i < 2 = g, with ri = ri+g = 1 and rj = 0 for all other indices j. Thus, V1 is similar to the irreducible 4-dimensional representation

ϕi ϕi+g ψ : PSL(2, q) → GL(4, Fq), A 7→ S1(A ) ⊗ S1(A ).

Lemma 4.9 shows that for any A ∈ PSL(2, q) the equality S1(A) = A holds; therefore ψ(A) = Aϕi ⊗ Aϕi+g .

35 4. Aﬃne polynomials with g ≤ 2

In the following we calculate for a given element A the dimension of the space of fixed points of ψ(A). Huppert [25, II 8.1] shows that the centralizer CPSL(2,q)(x) of any element A of order p is elementary abelian of order q. Hence, if A 6= 1, then either A has order p or is p-regular. 1 1 First let |A| = p. We may assume |A| = p. Then A is conjugate to ( 0 1 ) in GL(2, q). Thus, the dimension of the fixed point space of ψ(A) equals the dimension of the fixed 1 1 1 1 point space of ( 0 1 ) ⊗ ( 0 1 ) which is 2. Now let A be p-regular. Then A ∈ SL(2, q) is also p-regular and can be diagonalized −1 over a suitable extension field of Fq. Denote by λ, λ the eigenvalues of A. Due to Ortega [41, 6.3.1] eigenvalues of ψ(A) are exactly given by the product of eigenvalues of Aϕi with eigenvalues of Aϕi+g ; thus, ±1 ±1 λϕi λϕi+g and (λ−1)ϕi λϕi+g are the eigenvalues of ψ(A). We get

g i g (λϕi λϕi+g ) = 1 ⇐⇒ (λp +1)p = 1 ⇐⇒ λp +1 = 1 ⇐⇒ |A| | pg + 1 and

g i g (λ−1)ϕi λϕi+g = 1 ⇐⇒ (λp −1)p = 1 ⇐⇒ λp −1 = 1 ⇐⇒ |A| | pg − 1.

m Since g = 2 , we obtain  4 |A| = 1,  2 |A| = p, dim Fixψ(A) = √ √ Fq q−1 q+1 2 A 6= 1,A = 1, or A = 1,  0 otherwise.

As Vi and V1 are Galois conjugate, in both modules the dimension of the space of fixed q points of A is identical. Because N F is the direct sum of all Vi, h ∈ H fixes exactly one element of N unless h = 1; then h fixes q2 = n points,  √ |h| = p; then h fixes q = n points, √ √ √  q−1 q+1 h 6= 1 and |h| | 2 or |h| | 2 ; then h fixes q = n points. These improved estimations succeed in disproving the genus-0 condition: √ √ q+1 q+1 q+1 q−1 Suppose p ramifies tamely. As gcd( 2 , 2 ) = gcd( 2 , 2 ) = 1, 2 q + 1 ind(p) ≥ q2 − q2 + − 1 · 1 . q + 1 2 Together with the estimation for ind(∞) from Corollary 3.16 a violation of the genus-0 condition follows. If p ramifies wildly, then the equation 2 q(q − 1) 2 1 ind(p) ≥ q2 − q2 + − 1q + q2 − q2 + (q − 1)q q(q − 1) 2 q − 1 q holds. We obtain again a contradiction to the genus-0 condition.

36 4.1. g = 0 b = 1: We have to solve aJ < 4; this is only possible for (a, J) ∈ {(1, 2), (1, 3)}. Pm−1 Pg−1 The case (a, J) = (1, 2) cannot occur because 2 - i=0 ri = b i=0 Ri = 1.

In the other case ri = 2 for exactly one index i; all other rj vanish. We get m = g = f and b n = pfJ = p3m = q3.

V1 is similar to the irreducible 3-dimensional representation

a2 2ab b2ϕi a b ψ : PSL(2, q) 7→ GL(3, q), A = 7→ S (Aϕi ) = ac ad + bc bd . c d 2   c2 2cd d2

The dimension of the space of fixed points of ψ(A) can be determined as shown in the case b = 2. We get ( 3 A = 1, dim Fixψ(A) = (4.3) Fq 1 otherwise. √ Thus, any h ∈ H] fixes exactly 3 n = q points. These estimations are sufficient in disproving the genus-0 condition.

4.1.8. Case (H): H ∼= PGL(2, q) with p 6= 2 This case does not occur, either. We use the same methods as in the previous case. H is irreducible on N We use the notation of the corresponding section of case (G).

If q > 3, then H contains a unique proper and nontrivial normal subgroup. This group has index 2 in H, is characteristic and isomorphic to PSL(2, q). We obtain

∼ Gal Z|K(t) = M o F ∼ ∼ with F ∈ {1,C2, PGL(2, q)}. The cases F = C2 and F = 1 are impossible; thus, F = H. If q = 3, then H is isomorphic to S4. Every normal subgroup of H is also characteristic in H. Quotient groups of H are isomorphic to 1, S4, C2, or S3. Except for S4, all of these ∼ groups enforce M to be of dimension one. Thus, Gal Z|K(t) = M o PGL(2, 3). Bounds for the number of ﬁxed points for elements of H An analogue to Lemma 4.7 is

Lemma 4.12 Suppose 1 6= h ∈ H. Then

1 (1) dim Fix(h) ≤ 2 dim N if q > 2, |h|= 6 {2, 4}, and (|h|, q) 6= (3, 9),

2 (2) dim Fix(h) ≤ 3 dim N if (|h|, q) = (3, 9),

2 (3) dim Fix(h) ≤ 3 dim N if |h| = 4,

2 (4) dim Fix(h) ≤ 3 dim N if |h| = 2 and q 6∈ {3, 5}.

37 4. Aﬃne polynomials with g ≤ 2

Proof. The commutator subgroup H0 of H is a characteristic subgroup of H with H/H0 =∼ 0 ∼ 0 C2 and H = PSL(2, q). Thus, N considered as a H -module is semisimple. Let M denote an irreducible H0-submodule of N. Then either

N = M or N = M ⊕ M u with an appropriate u ∈ H.

H0 acts faithfully on M because the kernel of this operation is normal and, hence, characteristic in H0 and would be a subgroup of the kernel of the action of H on N. Lemma 4.7 shows that the bounds for ﬁxed point spaces of H0 are linear in the dimension of the module M; thus, these bounds also hold for the module N. Since h ∈ H ﬁxes less elements of N than h2 ∈ H0, we can use the bounds for h2 as bounds for h.

2 1 The assumption in case (1) gives |h | > 2. Therefore Lemma 4.6 yields dim Fix(h) ≤ 2 dim N. Case (2) can be proved analogously. Case (4) is a direct consequence from [21]. Case (3) follows from case (4) and some explicit calculations for q ∈ {3, 5}.

N is a “big” H-module: n ≥ q4 Suppose q 6∈ {3, 5, 9}. Then we get from Lemma 4.12  n h = 1,   n n | Fix(h)| ≤ 4 ≤ q |h| ∈ {2, 4},  q 3  n q2 otherwise. First we discuss the case where p ramiﬁes tamely with index q + 1. Then 1 ind(∞) ≥ (n − 1)1 + . q − 1

Ip is a cyclic group and contains therefore a unique involution and at most two elements of order 4. We get 1 n n ind(p) ≥ n − n + 3 4 + (q − 3) 2 . q + 1 q 3 q This gives

2 2 5 2 q (q + 1)(q − 2) + n(q − 3q 3 + 4q + 3q 3 − 3) ind(∞) + ind(p) − 2n + 2 ≥ ; q2(q2 − 1)

2 5 2 as q − 3q 3 + 4q + 3q 3 − 3 is positive for all possible q, the genus-0 condition is violated.

∼ Now suppose p ramiﬁes wildly. Then I∞ = N o Cq+1. By Lemma 4.8 Ip contains at most q involutions and 2q elements of order four. Using the same techniques as above gives again a contradiction.

The cases q ∈ {3, 5} can be disproved analogously; the number of ﬁxed points of elements n n that is not dealt with in Lemma 4.12 can be estimated with p = q . For q = 9 the same methods induce a contradiction to the genus-0 condition.

38 4.1. g = 0

N is a “small” module: n < q4 We use Theorem 4.11 to classify all possible H-modules N with n < q4. In this section the same notation is used as in the corresponding section of case (G).

V1 can be parametrized by V1 = (s; r0, . . . , rm−1).

The action of ξ on V1 gives

ξ V1 = (sp; rm−1, r0, . . . , rm−2).

f As ξ stabilizes V1, the parameters must fulﬁll

f sp ≡ s mod (q − 1) and ri = rj if i ≡ j mod g.

Thus, there exist elements R0,...,Rg−1 with 0 ≤ Ri < p such that

m−1 g−1 m X m X dim V = J g and r = R . Fq 1 i g i i=0 i=0

V1 being a faithful H-module gives J > 1; the equation

m dim N dim N Fq fJ g 4 b p Fp = p Fq = p < q ⇐⇒ aJ < 4b

enforces again b < 4. So, we have to discuss the following cases: b = 3: We obtain the unique solution (a, J) = (1, 2). This gives an index 0 ≤ i < g with Pm−1 i ri = ri+g = ri+2g = 1; all other rj vanish. As p is odd, the sum i=0 rip is also Pm−1 i odd. But the even number q − 1 cannot divide 2s + i=0 rip . This contradiction to Theorem 4.11 shows that this case is impossible. b = 2: We obtain the unique solution (a, J) = (1, 2). We further get f = g and 2g = m. The condition for s translates to

p2g − 1 | spg − s = s(pg − 1) ⇐⇒ pg + 1 | s;

g so, s = `·(p +1). We further get the existence of an index 0 ≤ i < g with ri = ri+g = 1 and rj = 0 for all other indices. The restriction from Theorem 4.11 shows

p2g − 1 | 2`(pg + 1) + pi + pi+g = (pg + 1)(2` + pi) ⇐⇒ pg − 1 | 2` + pi.

This is impossible since pg − 1 is even while 2` + pi is not. b = 1: In this case we have a = 1, m = f = g, and J ∈ {2, 3}. P i J = 2 is impossible for rip is odd. 3 So the remaining case is J = 3. We obtain n = q ; the action of PGL(2, q) on V1 is given by s ϕi ψ : PGL(2, q) → GL(3, q), A 7→ (det A) S2(A ). (4.4) s has to fulﬁll q − 1 q − 1 q − 1 | 2s + 2pi ⇐⇒ | s + pi ⇐⇒ s ≡ −pi mod ; 2 2

39 4. Aﬃne polynomials with g ≤ 2

as 0 ≤ s < q − 1, this congruence gives two different solutions for s. Note that the restriction of ψ to PSL(2, q) E PGL(2, q) coincides with the irreducible representation of the corresponding section of case (G). We use this fact to transfer results from equation (4.3) on page 37 to this case. An element of order p has a one-dimensional fixed point space because its image is 1 1 similar to ψ 0 1 . 2 Let A ∈ PGL(2, q) be a p-regular element of order 6= 2. Then 1 6= A ∈ PSL(2, q); equation (4.3) shows that the fixed point space of ψ(A) is at most one-dimensional. At last suppose A ∈ PGL(2, q) is an involution. As ψ is faithful, ψ(A) is at most two-dimensional. These bounds again induce contradictions to the genus-0 condition in all cases.

4.1.9. Case (I): H ∼= PGL(2, 2m) ∼ Suppose q = 2. Then H = D3 and the ramiﬁcation data is (2, 3). This situation has already been described in case (D).

Hence, we will assume q 6= 2 from now on. Then H is simple with H =∼ SL(2, 2m). By Lemma 4.4 N is an irreducible H-module. The “big” cases can be disproved essentially the same way as in case (G) and (H). The m “small” cases are more complicated. V1 is an irreducible PGL(2, 2 )-module induced by

m m s ϕi ψ1 : PGL(2, 2 ) → GL(2, 2 ), A 7→ (det A) · A , here n = q2; or

m m m s ϕi ϕi+ ψ2 : PGL(2, 2 ) → GL(4, 2 ), A 7→ (det A) · A ⊗ A 2 , 2 m here n = q and m is even with 0 ≤ i < 2 ; or m m m m s ϕi ϕi+ ϕi+2· ψ3 : PGL(2, 2 ) → GL(8, 2 ), A 7→ (det A) · A ⊗ A 3 ⊗ A 3

8 3 m here n = q and 3 | m with 0 ≤ i < 3 . In all cases s is uniquely determined; we have m det ψi(A) = 1 for all A ∈ PGL(2, 2 ). With the same methods as before it is possible to determine the dimension of the space of m ﬁxed points of the image of A ∈ PGL(2, 2 ) under ψi. We obtain  2 ψ1(A) = 1,  dim Fix ψ1(A) = 1 ψ1(A) is an involution, 0 otherwise;  4 ψ2(A) = 1,  √ √ dim Fix ψ2(A) = 2 |ψ2(A)| = 2 or |ψ2(A)| | q + 1 or |ψ2(A)| | q − 1 0 otherwise;  8 ψ3(A) = 1,  dim Fix ψ3(A) ≤ 4 ψ3(A) is an involution, 2 otherwise. These bounds imply a contradiction to the genus-0 condition in all cases.

40 4.2. g = 1

4.2. g = 1

In this section we suppose E to be a genus-one function field. As the field of constants of E is algebraically closed, it is the function field of an elliptic curve. We will prove

Theorem 4.13 With the notation from page 23 there is no aﬃne polynomial f such that g(E) = 1.

4.2.1. Ramiﬁcation in E|K(t)

Before we describe the structure of AutK (E) we remind the reader that the set of places P(E) of E can be canonically interpreted as an abelian group; we will denote its zero element by o. Hasse [23] shows that every K-automorphism of E acts on P(E) and, conversely, if any operation on P(E) induces an automorphism of E, then this automorphism is uniquely determined. In particular, the translation mappings

τa : P(E) → P(E), p 7→ p + a induce translation automorphisms in AutK (E). We denote the set of translation automorphisms by T; this set is closed under composition and forms a normal subgroup in AutK (E). Denote by I the stabilizer of o in AutK (E). Then I is a ﬁnite group of homomorphisms P(E) → P(E) and the set of K-automorphisms of E can be written as

AutK (E) = T o I. (4.5)

A ﬁrst consequence is

Lemma 4.14 Suppose σ = τa · ξ with τa ∈ T and ξ ∈ I ﬁxes a place P ∈ P(E). Then: (1) σ ∈ AutK (E) \ T ∪ 1.

σr ξr Pr ξi (2) For any Q ∈ P(E) and r ∈ N we have Q = Q + i=1 a . In particular, σ has order |ξ|.

Proof. σ ﬁxes P. No nontrivial translation automorphism has this property. This is (1).

The equation for Qσr follows by induction and the fact that elements of I are homomorphisms of the group P(E). |ξ| P|ξ| ξi P|ξ| ξi σ maps P to P + i=1 a . As every power of σ stabilizes P, it follows i=1 a = o. Thus, σ|ξ| = 1 and |ξ| is the smallest positive integer with this property.

The isomorphism types for I are determined in Husem¨oller[27, Chapter 3]; the following cases occur: I =∼ Conditions

C2 C4 p 6∈ {2, 3} C6 p 6∈ {2, 3} C3 o C4 p = 3, I non-abelian SL(2, 3) p = 2

41 4. Aﬃne polynomials with g ≤ 2

This allows us to work out how places can ramify in the extension E|K(t).

Lemma 4.15 Let p ∈ PK(t) be any place of K(t). Suppose P ∈ P(E) lies over p. Denote the inertia group of the extension P|p with Fp and the degree of the diﬀerent coming from the ramiﬁcation of p with δp. Set s resp. sk the number of groups in the series Fp(i) i≥1 of |Fp(1)| order 6= 1 resp. order pk−1 . Suppose |Fp| > 1. Then: ∼ ∼ I = Fp = δp = Conditions 1 C2 2 |H| p 6= 2 C2 1 C2 2 |H| · (1 + s) p = 2 1 C2 2 |H| C4 3 C4 4 |H| 1 C2 2 |H| 2 C6 C3 3 |H| 5 C6 6 |H| 1 C2 2 |H| 3 C4 4 |H| 1 C3 o C4 C3 3 |H| · (2 + 2s) 1 C6 6 |H| · (5 + 2s) 1 C3 o C4 12 |H| · (11 + 2s) 1 C2 2 |H| · (1 + s) 2 C3 3 |H| 1 C4 4 |H| · (3 + 3s1 + s2) s1 ≥ 1, s2 ≥ 1 SL(2, 3) 1 C6 6 |H| · (5 + s) 1 Q8 8 |H| · (7 + 7s1 + 3s2 + s3) s1 ≥ 1, s2 + s3 ≥ 1 1 SL(2, 3) 24 |H| · (23 + 7s1 + 3s2 + s3) s1 ≥ 1, s2 + s3 ≥ 1

Proof. The decomposition (4.5) of AutK (E) proves the existence of a group F ≤ I such that TFp = TF . By Lemma 4.14 Fp ∩ T = 1; thus, ∼ ∼ Fp = TFp/T = TF/T = F.

Hence, Fp embeds into I.

We obtain an analogue to Theorem 4.3:

Theorem 4.16 The following table gives the possible ramiﬁcation behavior in the extension E|K(t): Case I =∼ Ramiﬁcation Data Conditions (A1) (2,6) s = 2 C C (A2) 3 o 4 (4,12) s = 2 (B1) (2,6) s = 1 for both places (B2) (3,6) s = 3 (B3) (3,24) s = 1, s = 0, s = 2 SL(2, 3) 1 2 3 (B4) (6,6) s = 1 for both places (B5) (6) s = 7 (B6) (24) 25 = 7s1 + 3s2 + s3

42 4.2. g = 1

Proof. The proof is rather combinatorial; we use the genus formula of Riemann-Hurwitz together with the ramification data from Lemma 4.15 to ensure the correct genus of E. Some combinations, however, cannot occur although they formally satisfy the different- condition. This is the case if too many places ramify (Lemma 3.21), or more than one place ramifies and all ramification groups are p-groups (Corollary 3.16), or only ∞ ramifies, I∞ is a p-group, and the second ramification group of ∞ in the extension E|K(t) does not vanish (Corollary 3.11). These criteria exclude in particular all cases where I is a cyclic group. ∼ As an example we discuss the case I = C2. If p 6= 2 is odd, then the extension E|K(t) 1 is tame and the degree of the different of each branch point equals 2 |H|. Riemann-Hurwitz states 1 2g − 2 = 0 = −2|H| + a · |H| 2 with a being the number of branch points of E|K(t). Hence, a = 4; but this contradicts Lemma 3.21. Suppose p = 2. Lemma 3.18 shows that ∞ ramifies in the extension E|K(t); Corollary 3.16 gives F∞(0) = F∞(1) = C2 > F∞(2) = 1 and the nonexistence of any finite branch point. By Riemann-Hurwitz 1 2g − 2 = 0 = −2|H| + |H| · (1 + 1) = −|H| ⇐⇒ |H| = 0; 2 but this is impossible.

4.2.2. Cases in odd characteristic p > 2 We only disprove case (A2); the same methods also work for case (A1).

5 Suppose p ramiﬁes with index 12. Then ind(∞) ≥ 4 (n − 1) and 1 n 1 1 n 5 ind(p) ≥ n − n + 11 + 2 · n − n + 2 · = n. 12 3 4 3 3 6 But now ind(∞) + ind(p) > 2n − 2.

∼ Thus, Ip = C4. Section 3.2.1 gives 5 1 n ind(∞) ≥ (n − 1) + n − . 4 4 3 Furthermore 1 ind(p) = n − (n + f + 2f ) 4 2 4 where fi denotes the number of ﬁxed points of an element of order i in Ip. As ind(p) ∈ N, n n n the impossibility of f2 = 3 follows. Hence, f2 ≤ 9 and, thus, f4 ≤ f2 ≤ 9 . These bounds induce a contradiction to the genus-0 condition.

4.2.3. Cases in even characteristic p = 2

Supersingularity As AutP(E) =∼ SL(2, 3), by Silverman [46, V §3] E is the function ﬁeld of a supersingular elliptic curve. [46, V 3.1] gives some properties of such curves; for us most important is

43 4. Aﬃne polynomials with g ≤ 2

Lemma 4.17 There does not exist p ∈ P(E) with p 6= o and 2 · p = o.

For the following we identify H with the Galois group of E|K(t). Then, a ﬁrst consequence of the above lemma is

Lemma 4.18 Suppose h ∈ H is an involution. Then h ﬁxes either no place or exactly one place of E.

Proof. h can be written as a product h = τξ with τ ∈ T and ξ ∈ I. If ξ = 1, then the assertion follows immediately. Thus, by Lemma 4.14 ξ 6= 1 is the unique involution in I =∼ SL(2, 3). ξ acts on P(E) by inversion, i.e. every place p of E is mapped to −p. Suppose h ﬁxes the places p, q ∈ P(E). Then ph − p = o = qh − q ⇐⇒ pξ − p = qξ − q ⇐⇒ 2 · (p − q) = o and we get p = q by Lemma 4.17.

Now we are able to state the main observation of this section

Proposition 4.19 Let i ∈ H be an involution that ﬁxes the place P ∈ P(E). Then there exists a bijection between the set iH of H-conjugates of i and the set PH of places over p := P ∩ K(t). In particular, the normalizer NH (hii) of hii in H equals the inertia group Fp of P|p. ∼ Proof. As Fp embeds into I = SL(2, 3), i is the unique involution in Fp. Therefore hii is normal in Fp; hence, Fp ≤ NH (hii). Let Q ∈ PH be another place lying over p. Then the inertia group of Q|p is conjugate to H H Fp and, thus, contains exactly one involution from the set i . Since every involution in i ﬁxes exactly one place of E, the equality |iH | = |PH | holds. The equation |H| |H| |iH | = = = |PH | |NH (hii)| |Fp| shows |NH (hii)| = |Fp|.

Corollary 4.20 In cases (B3) and (B6) a 2-Sylow subgroup of H is isomorphic to a quaternion group of order 8. In the remaining cases a 2-Sylow subgroup of H is isomorphic to C2.

Proof. We start with cases (B3) and (B6). Let F ≤ H be an inertia group of order 24, S ≤ F the unique 2-Sylow subgroup of F , and Z := Z(F ) ≤ S the center of F . Proposition 4.19 shows

F ≤ NH (S) ≤ NH (Z) = F ; as 2 - [NH (S): S], S is a 2-Sylow subgroup of H. Since F is isomorphic to SL(2, 3), the isomorphism type of S follows for instance from [32, 8.6.10]. The remaining cases can be proved similarly.

44 4.2. g = 1

Cases (B1) and (B4) In both cases a 2-Sylow subgroup of H has order 2; thus, the set of involutions of H forms a conjugacy class of H. By Theorem 4.16 two diﬀerent places of K(t) ramify with even index; therefore an involution of H ﬁxes more than one place of E. This contradicts Lemma 4.18. Case (B2) This case can be disproved by using the estimations for ind(∞) and ind(p). Case (B3) This case does not occur.

∼ The estimations for ind(∞) and ind(p) show the impossibility of Ip = SL(2, 3). Hence,

4 2 n ind(∞) ≥ (n − 1) + 2 · n − 3 24 2

1 and ind(p) = n − 3 (n + 2f3) with f3 being the number of ﬁxed points of a generator of the n cyclic group Ip. The genus-0 condition and ind(p) being an integer give f3 = 4 .

A more detailed description of H is given by

Lemma 4.21 H contains a characteristic and abelian subgroup J such that H = J oF∞ and h J ∩ Fp = 1. Let Q ≤ F∞ denote a 2-Sylow subgroup of H. Then F := hQ | h ∈ Hi = J o Q is a Frobenius group with Frobenius kernel J and complement Q; [H : F ] = 3. F is also characteristic in H.

Proof. Let J denote a maximal normal subgroup of H of odd order. As Q is isomorphic to a quaternion group of order 8, a theorem of Brauer/Suzuki ([25, V 22.9] or [4]) states that the center of H/J is cyclic of order two. Therefore H contains another normal subgroup J˜ with [J˜ : J] = 2.

Remember our identification of H with the Galois group of E|K(t) and set Y := Fix(J), Z := Fix(J˜). Suppose the extension E|Y is ramified. Then g(Y ) = 0 and Y is a rational function field. Since J has odd order, the inertia group of a place of Y lying over ∞ contains a quaternion group of order 8 in contradiction to the classification [50] of AutK Y |K(t) . Hence, E|Y is unramified with g(Y ) = 1; this shows in particular that J ∩ Fp = 1. Fur- thermore by [46, III §4] we may assume J to be the set of automorphisms of E coming from the kernel of an isogeny E → Y ; thus, J is abelian. As 8 - [H : J˜], the extension Y |Z is ramified; every branch point of Z ramifies totally with index 2 and adds 4 to the degree of the different of Y |Z. Hence, exactly one place ramifies in Y |Z. As this place lies over ∞, the ∼ classification [50] of AutK Z|K(t) yields Gal Z|K(t) = A4. We obtain that ∞ ramifies ∼ ∼ totally in the extension Y |K(t) with Gal Y |K(t) = SL(2, 3) = F∞. By Herbrand F∞ is a complement of J in H.

Proposition 4.19 shows that J is regular on the set of H-conjugates of Q (by conjugation); hence, F ≤ J o Q. Since F is transitive on its 2-Sylow subgroups, the equality F = J o Q holds. As J > 1 by Proposition 2.8, the remaining assertions follow as Q is selfnormalizing in F .

45 4. Aﬃne polynomials with g ≤ 2

Next we classify all absolutely irreducible F2-modules of F . Our main tool for this is the representation theory of Frobenius groups. The following proposition states the results for the convenience of the reader. A short and elegant proof can be found in Guralnick [16].

Proposition 4.22 Let F = J o Q be a Frobenius group with Frobenius kernel J and complement Q. Suppose K to be an algebraically closed ﬁeld. (1) Q acts semiregularly on the set of isomorphism classes of nontrivial irreducible K[J]- modules.

(2) If V is an irreducible K[F ]-module, then either CV (J) = V or there exists an absolutely irreducible nontrivial K[J]-module W such that V =∼ W F is isomorphic to the induced K[F ]-module W F . We use this proposition to get

Proposition 4.23 Let V be an irreducible H-submodule of N K := N ⊗ K. Then V is 8-dimensional. If t ∈ Fp has order 3, then the space of ﬁxed points of t on V is at most 4-dimensional. Proof. Let Q, F , and J be deﬁned as in Lemma 4.21. Ls Since F EH, the F -module V is semisimple and can be written as a direct sum V = i=1 Vi of irreducible F -modules 1 6= Vi ≤ V . We use Proposition 4.22 to clarify the structure of

V1: As J is normal in U, the case CV1 (J) = V1 is impossible. Hence, V1 is isomorphic to an absolutely irreducible and nontrivial J-module W induced to F . The commutativity of J gives dimK W = 1; since [F : J] = 8, it follows dimK V1 = 8. Because hF, ti = H, the group Fp is transitive on the Vi; therefore either V = V1 is an t t2 irreducible F -module or V = V1 ⊕ V1 ⊕ V1 . We show that the second case is impossible. V1 t intersects the space Fix(t) of ﬁxed points of t trivially for otherwise V1 ∩ V1 6= 0. But then the dimension formula for vector spaces gives dimK Fix(t) + dimK V1 = dimK Fix(t) + V1 + dimK Fix(t) ∩ V1 ; | {z } | {z } | {z } =r−2 ≤r =0 hence, the contradiction dimK V1 ≤ 2 follows. This gives dimK V = 8.

By Lemma 4.21 t can be written as a product t = jw with j ∈ J and w ∈ F∞ being of order 3. As the abelian group J is normal in H, the J-module V is a direct sum of eight one-dimensional J-submodules M1,...,M8 that are permuted transitively by F∞. Hence, Q E F∞ acts regularly on the modules Mi. ] Suppose n ∈ N generates M1. By proper identiﬁcation of the integers 1,..., 8 with the modules Mi the action of w on the set {M1,...,M8} is described by the permutation ω := (1)(2)(3, 4, 5)(6, 7, 8). Set qi the well-deﬁned element of Q that maps M1 to the module belonging to the integer i. Then 8 M V = hnqi i and hnqi iw = hnqiω i. i=1

] qi jw qiω P8 qi It follows that there exist elements ci ∈ K with (n ) = ci·n . Suppose v := i=1 λin ∈ V . Then 8 8 t jw X qiω X qi v = v = λicin = λiω−1 ciω−1 n . i=1 i=1

46 4.3. g = 2

Hence, v is ﬁxed by t if and only if λi = λiω−1 ciω−1 for all 1 ≤ i ≤ 8. Solving these equations shows that Fix(t) is at most 4-dimensional.

r The above proposition gives r = 8s with s ∈ N and dim Fix(t) ≤ 2 ; thus, dim Fix(t) = r − 2 is impossible. Case (B5) By Corollary 4.20 the order of H is 2u with 2 - u. Therefore H contains a normal subgroup J of index 2. In the extension E| Fix(J) exactly one place ramiﬁes tamely with index 3. But this gives a contradiction to the Riemann-Hurwitz genus formula. Case (B6) This case is impossible, too.

The idea of Lemma 4.21 can be used to obtain a contradiction in this case. With the notation of this lemma we get eventually that Z|K(t) is an Galois extension of rational function fields with the unique branch point ∞. The classification of Valentini and Madan [50] shows that the ramification index of ∞ in Z|K(t) has to be a power of 2. This, however, is impossible as E|Y is unramified.

Remark 4.24 This case can also be disproved without using the theorem of Brauer-Suzuki: A discussion of the ramification behavior of E| Fix(F∞) shows that in this extension exactly one place ramifies. Hence, the intersection of F∞ with any different conjugate is trivial. Since F∞ is selfnormalizing in H, the group H is a Frobenius group with Frobenius complement F∞. Now, the same considerations as above lead to a contradiction. ?

4.3. g = 2

In this section we assume E to be a function field of genus two. Schmid [42] shows that AutK (E) is a finite group. Let F denote the fixed field of all K-automorphisms of E. Then E|F is Galois with Gal(E|F ) = AutK (E). Stichtenoth [47, VI.2] shows that E is the function field of a hyperelliptic curve. Thus, E|F contains a unique intermediate rational field R with [E : R] = 2. The uniqueness of R shows that the extension R|F is Galois. The central involution ı ∈ AutK (E) that corresponds to R by Galois duality is called the hyperelliptic involution of E. We will prove

Theorem 4.25 With the notation from page 23 suppose g = 2. Then p = 3 and A = G = AGL(2, 3). f is not exceptional. Examples of f are given by the AGL-polynomials of chapter 5.

4.3.1. Cases in odd characteristic p > 2

The K-automorphisms for genus-two hyperelliptic function ﬁelds are well-known. Due to Geyer [15], Igusa [28], or Shaska/V¨olklein [45] only the following isomorphism types can occur:

47 4. Aﬃne polynomials with g ≤ 2

∼ AutK (E) = Conditions C2 C10 p 6= 5 D2 D4 D6 C3 o D4 p 6∈ {3, 5}, AutK (E) non-abelian GL(2, 3) p 6= 5 Σ5 p = 5

1 Here Σ5 denotes a non-split extension of C2 by S5 where the transpositions of S5 lift to involutions, cf. [45]. Σ5 is unique up to isomorphism. Lemma 4.1 and our main assumption p 6= 2 show that AutK (E) cannot be isomorphic to C2, C10, D2, D4, or C3 o D4. 0 As a subgroup of D6 either is a 3 -group or contains a characteristic subgroup isomorphic ∼ to C3, Proposition 2.8 gives the impossibility of AutK (E) = D6.

∼ Case AutK (E) = GL(2, 3) ∼ Assume AutK (E) = GL(2, 3); then p = 3. Lemma 4.1, Proposition 2.9, and Proposition 2.8 force H to be isomorphic to SL(2, 3) or GL(2, 3). A survey of the subgroups of these groups gives

Lemma 4.26 Let p ∈ PK(t) be any place of K(t). Suppose P ∈ P(E) lies over p. Denote the inertia group of the extension P|p with Fp and the degree of the different coming from the ramification of p with δp. Denote by s the well-defined integer with Fp(s) > Fp(s + 1) = 1. Suppose |Fp| > 1. Then: ∼ ∼ H = Fp = δp 1 C2 2 |H| 3 C4 4 |H| SL(2, 3) 1 C3 3 |H| · (2 + 2s) 1 C6 6 |H| · (5 + 2s) 1 C2 2 |H| 3 C4 4 |H| 7 GL(2, 3) C8 8 |H| 1 C3 3 |H| · (2 + 2s) 1 C6 or D3 6 |H| · (5 + 2s) We use this lemma to solve the Riemann-Hurwitz equation X 2 = −2|H| + δp p where p runs through all places of K(t) ramifying in E|K(t). We obtain the following

Theorem 4.27 The following table gives the possible ramiﬁcation behavior in the extension E|K(t):

1MAGMA [3] knows this group as “SmallGroup(240,90)”

48 4.3. g = 2

Case H =∼ Ramiﬁcation Data Conditions (A) SL(2, 3) (4,3) s = 1 (B) GL(2, 3) (8,6) s = 1

Case (A) can be easily disproved in the same manner as in the previous chapters. Case (B), however, is more interesting:

First it is easy to see that p must ramify with index 6. An inspection of the subgroups of H of order 8 shows that F∞ contains the hyperelliptic 48 48 involution ı. Thus, ı 6∈ Fp because otherwise 8 + 6 = 6 + 8 = 14 places would ramify with index 2 in the extension E| Fix(hıi), contrary to our assumption g = 2. ∼ As all subgroups of H of order 6 not containing ı are isomorphic to S3, we get Ip = S3. Theorem 4.27 shows that Ip(2) = 1; thus, 1 1 1 7n − 4f − 3f ind(p) = n − (n + 3f + 2f ) + n − (n + 2f ) = 3 2 6 2 3 2 3 3 6 where f2 resp. f3 is the number of ﬁxed points of an element of order 2 resp. 3 of Ip. 9 Since ind(∞) ≥ 8 (n − 1), a simple calculation shows that the genus-0 condition is violated n n 7 if f2 6= 3 6= f3. It follows f2 = f3 = 3 and ind(p) = 9 n.

Since I∞(1) ≤ N, section 3.2.1 gives ∞ ∞ 9 X 1 n 9 1 X 1 ind(∞) = (n − 1) + n − = (n − 1) + n 1 − . 8 8 |I (u )| 8 8 |I (u )| i=2 ∞ i i=2 ∞ i

The genus-0 condition forces I∞(u3) = 1. Therefore 7 9 1 1 63 |I (u )| n + (n − 1) + n1 − = 2n − 2 ⇐⇒ n = ∞ 2 . 9 8 8 |I∞(u2)| 9 − 2 |I∞(u2)| 2 Thus, I∞(u2) = 1, n = 9, and ind(p) = 3 − 2 = 7.

9 8 6 The above situation is realized for instance by f(X) = X + X + X ∈ F3[X], cf. chapter 5. As G = AGL(2, 3) is 2-transitive on N, this case never gives an exceptional polynomial.

∼ Case AutK (E) = Σ5

An inspection of the subgroups and the conjugacy classes of Σ5 shows that in case of tame ramiﬁcation inertia groups are isomorphic to

C2,C3,C4,C6, or C8, in case of wild ramiﬁcation to

C5,C10,C5 o C4, or C5 o C8. Since H may not contain a characteristic 5-Sylow subgroup by Proposition 2.8, it follows that H is isomorphic to either SL(2, 5) or Σ5. As Σ5 contains only one central involution, this involution coincides with the hyperelliptic involution ı. Note that ı is the only involution in case H =∼ SL(2, 5).

It follows

49 4. Aﬃne polynomials with g ≤ 2

Lemma 4.28 Let p ∈ PK(t) be any place of K(t). Suppose P ∈ P(E) lies over p. Denote the inertia group of the extension P|p with Fp and the degree of the different coming from the ramification of p with δp. Denote by s the well-defined integer with Fp(s) > Fp(s + 1) = 1. Suppose |Fp| > 1. Then: ∼ ∼ Fp = can occur for H = δp = 1 C2 all cases 2 |H| 2 C3 all cases 3 |H| 3 C4 all cases 4 |H| 5 C6 all cases 6 |H| 7 C8 Σ5 8 |H| 1 C5 all cases 5 |H| · (4 + 4s) 1 C10 all cases 10 |H| · (9 + 4s) 1 C5 o C4 all cases 20 |H| · (19 + 4s) 1 C5 o C8 Σ5 40 |H| · (39 + 4s) We use this lemma to solve the Riemann-Hurwitz equation X 2 = −2|H| + δp p where p runs through all places of K(t) ramifying in E|K(t). We obtain

Theorem 4.29 The following table gives the possible ramiﬁcation behavior in the extension E|K(t):

Case H =∼ Ramiﬁcation Data Conditions (A) SL(2, 5) (3,20) s = 2

(B) Σ5 (6,40) s = 2 ∼ In case (A) p cannot ramify with index 20; thus, we have Ip = C3 and 1 ind(p) = n − (n + 2f) 3 with f the number of ﬁxed points of an element of order 3 in Ip. ind(p) being an integer n 16 enforces f ≤ 25 . Therefore ind(p) ≥ 25 n. As 5 5 n ind(∞) ≥ (n − 1) + n − , 4 20 5 the genus-0 condition is violated.

∼ In case (B) p cannot ramify with index 40. Thus, we have Ip = C6 and 9 5 n ind(∞) ≥ (n − 1) + n − . 8 40 5

Let σ ∈ Ip be a generator for Ip and denote by A ∈ GL(r, 5) the automorphism of N that is induced by the action of σ ∈ H on the F5-vector space N; both σ and A have order 6. As 6 F25 is a splitting ﬁeld for the separable polynomial X − 1 ∈ F5[X], A is diagonalizable as 0 an automorphism of N := N ⊗ F25.

50 4.3. g = 2

First suppose that a primitive sixth root of unity is an eigenvalue of A. Then every Galois conjugate of this root of unity is also an eigenvalue of A; hence, A has two sixth roots of s s n unity in its spectrum. Thus, every power A of A with A 6= 1 ﬁxes at most 25 points. We obtain 1 n 4 ind(p) ≥ n − n + 5 = n. 6 25 5 This violates the genus-0 condition with the above estimation for ind(∞). Next suppose that no primitive sixth root of unity is an eigenvalue of A. Then A has both a primitive third root of unity and −1 ∈ F5 in its spectrum. Again the inverse of the third 5 n 2 root of unity is also an eigenvalue of A. This shows that A and A have at most 125 , A 4 n 3 n and A at most 25 , and A at most 5 ﬁxed points. We get 1 n n n 98 ind(p) ≥ n − n + 2 + 2 + = · n. 6 125 25 5 125 This is impossible, too.

4.3.2. Cases in even characteristic p = 2 Geyer [15] classiﬁes all possible K-automorphism groups of genus-2 function ﬁelds; a nice list is also given in [8]. We obtain that AutK (E) is isomorphic to

C2,C2 × C2,D6,G32, or G160 where Gi is some group of order i. We will describe the groups Gi later when we need information about their internal structure.

∼ Proposition 2.9 (2) reduces considerably the work in this section: only the cases AutK (E) = ∼ D6 or AutK (E) = G160 may occur.

∼ Case AutK (E) = D6 ∼ ∼ By Lemma 4.1 and Proposition 2.8 it follows H = D3 as Z(D6) = C2. Riemann-Hurwitz and Corollary 3.16 imply the impossibility of this case.

∼ Case AutK (E) = G160

First we have to get some information about the group G160. Due to [8, Sections 2.1, 3.1] G160 sits in the middle of the non-split exact sequence

4 1 −→ hıi −→ G160 −→ C2 o C5 −→ 1

4 where C5 acts non-trivially on C2 . 2 Up to isomorphism G160 is uniquely determined by this sequence; we get

∼ − G160 = E32 o C5

− where E32 denotes the extraspecial group of order 32 being a central product of a quaternion group of order 8 and two copies of the dihedral group of order 8 (cf. [25, III 13.8]) and C5

2MAGMA [3] knows this group as “SmallGroup(160,199)”

51 4. Aﬃne polynomials with g ≤ 2

− acts nontrivially on E32. ∼ An inspection of the subgroups of G160 and Proposition 2.9 prove that either H = C10 or ∼ H = G160. Hence, in every case H contains a characteristic nontrivial 2-subgroup. This contradicts Proposition 2.8.

52 5. AGL as a monodromy group

In this chapter f denotes a polynomial of the form

s pr X pr−pi f(X) := X + aiX with s < r and a0as 6= 0. (5.1) i=0 We are interested in the calculation of the geometric monodromy group of f. Abhyankar [1] already dealt with the case s = 0; he proved that f is aﬃne with G = AGL(1, pr). Thus, for the rest of this chapter we will always assume s 6= 0; this implies in particular r ≥ 2. We use the notation from page 23.

Lemma 5.1 (1) f(X) − t is separable.

(2) f is functionally indecomposable over K.

(3) G ≤ AGL(r, p). (We do not prove here that G is an aﬃne group.) Proof.

0 pr−2 (1) The derivative of f is given by f (X) = −a0X 6= 0. Therefore f(X) − t does not have multiple roots.

(2) Suppose there exist nonlinear polynomials g, h ∈ K[X] such that f = g ◦ h. We may assume h(0) = 0. Since f(0) = 0, this yields g(0) = 0, too. It follows from (1) that 0 0 0 pr−2 f (X) = g h(X) · h (X) = −a0X . 0 0 pr−2 0 r Suppose g (0) 6= 0. Then X - g ◦ h and we get X |h . Hence, deg h ≥ p − 1. But then pr = deg f = deg g · deg h ≥ 2(pr − 1) which is impossible. Thus, g0(0) = 0. As 0 ∈ K is the only zero of f 0, h(ξ) 6= 0 for all ξ ∈ K]. We obtain α r h(X) = h0X . But then every summand of f has degree divisible by α. Since p and pr − 1 are relatively prime, this enforces α = 1 contrary to our assumption.

−1 −1 (3) Let x1, . . . , xn ∈ L be the roots of f(X) − t. Set Z := X and zi := xi ; this is well- deﬁned as all xi are diﬀerent from zero. An easy calculation shows that the equation f(X) − t = 0 can be rewritten in the form

s r X 1 i 1 Zp − a Zp = . (5.2) t i t i=0

Since the zeros of this equation are precisely the elements zi ∈ L, the splitting ﬁeld of (5.2) is L. Section 2.3 gives the claim.

The following concept severely restricts the possibilities for G; we will prove for instance that G is 2-transitive.

53 5. AGL as a monodromy group

Definition 5.2 (Jordan group) Let G be a group acting transitively on the finite set Ω. A subset Γ ⊆ Ω is said to be a Jordan set if |Γ| > 1 and the pointwise stabilizer G(Ω\Γ) is transitive on Γ. The set ∆ := Ω \ Γ is called a Jordan complement. If G is k-transitive on Ω, every subset ∆ of size < k is a Jordan complement; in such a case we call Γ and ∆ improper, otherwise we call them proper. If Γ = Ω, we call Γ and ∆ trivial. G is called a Jordan group if it has at least one proper Jordan complement. In our situation this definition yields

Proposition 5.3 G is a 2-transitive group and contains a nontrivial Jordan complement ∆ of size |∆| = ps. If p 6= 2 or p = 2 and s > 1, then ∆ is proper and G is a Jordan group. Proof. We consider the ramiﬁcation of the place 0 : t 7→ 0 in the extension K(x)|K(t). Since s pr−ps ps X ps−pi t = f(X) = X · X + aiX i=0 ps Ps ps−pi with X + i=0 aiX being a separable polynomial, 0 decomposes over K(x) in the following way:

pr−ps 0 = P0 · P1 ··· Pps with pairwise diﬀerent places Pi ∈ P K(x) . (5.3)

s Hence, by van der Waerden [51] Ip fixes a set ∆ of p elements pointwise and permutes the remaining pr −ps elements transitively. Denote this orbit by Γ. Then the pointwise stabilizer G(∆) of ∆ in G contains Ip; thus, G(∆) is a fortiori transitive on Γ. Γ is a nontrivial Jordan subset of G. The indecomposability of f implies the primitivity of G; hence, Neumann [40, Theorem J1] gives the 2-transitivity of G. Suppose p > 2. Lemma 5.1 and the classification in [32, 4.2.5] show that G is 2-transitive but not 3-transitive. Hence, ∆ being proper comes down to |∆| ≥ 2; but this condition is always fulfilled. Suppose p = 2. [32, 4.2.5] shows that G is at most 4-transitive. The claim follows now easily.

Remark 5.4 The 2-transitivity of G can be obtained in a diﬀerent way, too. Unfortunately, the following proof does not show the important fact that G is a Jordan group in almost all cases. Let X and Y be algebraically independent transcendentals over K. Set φ(X,Y ) := f(X)− φ(X,Y ) f(Y ) ∈ K[X,Y ]. We prove that X−Y is absolutely irreducible. Suppose φ(X,Y ) = Ak(X,Y ) + Ak−1(X,Y ) + ... Bm(X,Y ) + Bm−1(X,Y ) + ... with Ai,Bi ∈ K[X,Y ] being homogeneous polynomials of degree i. Then

pr pr pr AkBm = X − Y = (X − Y ) ;

k m this gives Ak = (X − Y ) and Bm = (X − Y ) . Hence,

pr−1 pr−1 k m a0 X − Y = AkBm−1 + Ak−1Bm = (X − Y ) Bm−1 + (X − Y ) Ak−1.

54 As the left hand side of this equation is separable, it follows k = 1 or m = 1. The 2-transitivity of G follows for instance from [14, Exceptionality Lemma]. ?

For the following we need some information about the p-structure of GL(r, p).

Lemma 5.5 GL(r, p) does not contain an element of order pr. GL(r, p) contains an element of order pr−1 if and only if r ∈ {1, 2} or (p, r) = (2, 3).

Proof. Let P ≤ GL(r, p) be a p-Sylow subgroup of GL(r, p). Denote by Js a Jordan block of dimension s for the eigenvalue 1. The minimal polynomial of Js is given by µs(X) := s (X −1) . Every A ∈ P is similar to a direct sum of Jordan blocks Jsi , i.e. we ﬁnd an element T ∈ GL(r, p) with a a T M X A = Jsi and si = r. i=1 i=1

Let s := max{si | 1 ≤ i ≤ a} denote the maximum of the si. Then the minimal polynomial i of A is given by µs. The order A is given by p where i is the smallest integer such that pi µs | (X − 1) . This proves that Jr has the highest possible p-order in GL(r, p). We have

i i−1 i |Jr| = p with p < r ≤ p .

As the equation pr−1 < r is not solvable, GL(r, p) never contains an element of order pr. It follows by induction that the equation pr−2 < r holds exactly for r ∈ {1, 2} or (p, r) = (2, 3).

Corollary 5.6 G is an aﬃne group. Moreover, there exists a divisor e of r such that either r r ∼ ASL(e, p e ) ≤ G ≤ AΓL(e, p e ) or N o A7 = G ≤ AGL(4, 2).

4 In the latter case A7 ,→ GL(4, 2) acts 2-transitively on the nonzero elements of F2. Proof. Let S := soc(G) be the socle of G. As G is 2-transitive, S is either elementary abelian and regular or isomorphic to a non-abelian simple group.

We start with the latter case. Suppose S is non-abelian and simple.

We show first that S can only be isomorphic to a projective special linear group. Assume G is a Jordan group. The primitivity of G and Neumann [40, Classification Theorem] immediately show that S =∼ PSL(d, q) acts 2-transitively on the projective plane qd−1 r PG(d − 1, q) with | PG(d − 1, q)| = q−1 = p . If G is not a Jordan group, then by Proposition 5.3 (p, s) = (2, 1) and – with the notation from this proposition – the two-point stabilizer G(∆) is transitive on Γ. Hence G is even 3-transitive. The classification of 2-transitive groups (a nice list is given in Cameron [6]) shows that soc(G) is isomorphic to PSL(2, q) acting on PG(1, q) with | PG(1, q)| = q +1 = 2r.

Suppose S =∼ PSL(d, q). We prove that this implies (pr, d, q) = (8, 2, 7). S is a subgroup of the aﬃne group AGL(r, p). Denote the aﬃne kernel of AGL(r, p) with N. The simplicity of S gives S ∩ N = 1. Hence, a point stabilizer of the group N o S is

55 5. AGL as a monodromy group isomorphic to S. Thus, S can be considered a subgroup of GL(r, p). The integers pr, d, and q fulﬁll the equation qd − 1 pr = . (5.4) q − 1

First assume that qd − 1 has a primitive prime factor. In our case as a consequence of Zsigmondy [52] this is true if d > 2, or d = 2 and q is even. By (5.4) this prime factor qd−1 is uniquely given by p; in particular, q−1 and (q − 1) are relatively prime. Let x be a qd−1 Singer-element of PGL(d, q). Then x has order q−1 , cf. Huppert [25, II 7.3]. As PSL(d, q) E q−1 PGL(d, q) with PGL(d, q)/ PSL(d, q) ≤ Cq−1 cyclic, x is an element of PSL(d, q) with |x| = |xq−1| = pr. We obtain that GL(r, p) contains an element of order pr. This contradicts Lemma 5.5. Now assume d = 2 with odd q. Then p = 2 is even. The same idea as above shows x2 ∈ PSL(2, q). Hence, GL(r, 2) contains an element of order 2r−1. Lemma 5.5 and Huppert [25, II 6.14] show that this is only possible for r = 3. Thus, we obtain (pr, d, q) = (8, 2, 7). But this is impossible: As S is selfnormalizing in AGL(3, 2), we obtain S = G. An explicit calculation shows that the Jordan complements of S have order 1 or 7 contrary to Proposi- tion 5.3.

Thus, we end up in the aﬃne case. If G is a Jordan group, Neumann [40] gives the assumption. Otherwise G is a 3-transitive aﬃne group in even characteristic. Our claim follows directly from Cameron/Kantor [7, Sec. 8].

Now we are able to state our main theorem:

Theorem 5.7 Deﬁne e the greatest common divisor of r and all 0 ≤ i ≤ s with ai 6= 0, r e e := gcd(r, i | ai 6= 0). Then G = AGL( e , p ).

Proof. We use the notation from Lemma 5.1 and Lemma 2.12.

Let v ∈ V ] be a nonzero element of V . Then the equation

s r X 1 i vp −1 − a vp −1 = 0 t i i=0 holds; in fact, the set of zeros of this equation coincides with V ]. Using the identity Z = X−1 we obtain s s r X 1 i X r i Zp −1 − a Zp −1 = 0 ⇐⇒ a Xp −p − t = 0. t i i i=0 i=0

Since the latter polynomial is irreducible, we see that the elements 1, v, v2, . . . , vpr−2 are linearly independent over K. Thus, any representation of vpr−1 by smaller powers of v is Ps 1 pi−1 uniquely given by i=0 t aiv .

56 Let c be an element of K×. Then the following holds:

s s r r X 1 i i r X 1 i cv ∈ V ] ⇐⇒ cp −1vp −1 = a cp −1vp −1 = cp −1 a vp −1 t i t i i=0 i=0 pr pi ⇐⇒ aic = aic for all 0 ≤ i ≤ s pr pi ⇐⇒ c = c for all 0 ≤ i ≤ s with ai 6= 0 a06=0 pr pi ⇐⇒ c = c ∧ c = c for all 1 ≤ i ≤ s with ai 6= 0 \ ⇐⇒ c ∈ Fpr ∩ Fpi = Fpe ≤ K. ai6=0,i6=0

r e Thus, we can consider V to be an Fpe -vector space and H ≤ ΓL( e , p ) is a subgroup of r e h h ΓL( e , p ). But H acts as a proper linear group since (cv) = cv for all h ∈ H and r e c ∈ Fpe ≤ K. Hence, H ≤ GL( e , p ).

r e In the following we prove that the equality H = GL( e , p ) holds. By Lemma 2.12 N equals the kernel of the operation of G on V . Therefore we have E = K(V ); but we can also think of E as the splitting field of the irreducible polynomial Ps pr−pi F (X) := i=0 aiX − t. Let v ∈ E be any root of F . Denote by F∞ the inertia group of a fixed place of E lying over ∞. The structure of F shows that ∞ ramifies totally in the extension K(v)|K(t). Hence, by ] van der Waerden [51] F∞ is transitive on V . Let P be the normal p-Sylow subgroup of F∞. As the p-group P fixes at least one element of V ], P fixes every element of V ] because all P -orbits on V ] have the same length. We 0 obtain P = 1; F∞ is a cyclic and transitive p -group of linear transformations of V of order r a multiple of deg F = p − 1. It follows from Huppert [25, II 3.10 and 7.3] that F∞ is a full r e Singer cycle in GL( e , p ). ∼ Using the classification from Corollary 5.6 we see that N o A7 = G ≤ AGL(4, 2) is im- r e possible as this group does not contain a cyclic subgroup of order 15. Hence, SL( e , p ) is a subgroup of H. By Huppert [25, II 7.3 (b)]

r e r e r e r e r e H/ SL , p ≥ SL , p F / SL( , p ) =∼ C e =∼ GL( , p )/ SL( , p ); e e ∞ e p −1 e e

r e r e as H ≤ GL( e , p ), it follows H = GL( e , p ).

Higher ramiﬁcation in L|K(t) Proposition 5.8 The normal p-Sylow subgroup of I∞ equals N. Moreover

ind(∞) = n and I∞(2) = 1.

s I0 is isomorphic to P o Cpr−s−1 with |P | = p . Moreover

ind(0) = n − 2 and I0(2) = 1. ∼ Proof. Theorem 5.7 gives I∞ = N o Cn−1; therefore we have 1 ind(∞) ≥ (n − 1) 1 + = n and ind(∞) = n ⇐⇒ I (2) = 1. n − 1 ∞

57 5. AGL as a monodromy group

We use the notation from the proof of Proposition 5.3. Let P be a place of L lying over P0. Denote by I the inertia group of the extension P|0. We know that I is isomorphic to a 0 semidirect product P o C with a p-group P and a cyclic p -group C. By van der Waerden [51] we obtain o(I) = s + 1. Denote by Γ the I-orbit consisting of r s p −p elements. As I(1) is normal in I, Γ splits into a diﬀerent I(1)-orbits γi, each of length b. Since C acts transitively on the γi, it follows

a | |C| and b | |P |. (5.5)

Because ab = |Γ| = ps(pr−s − 1), we get at once a = pr−s − 1, b = ps, and oI(1) = ps + pr−s − 1. Next we show that |I| = pr − ps. Suppose |I| > pr − ps. Then there exists an element 1 6= h ∈ I ∩ H. Since the inertia group of any place of L over 0 is G-conjugate to I, the G T g decomposition (5.3) of 0 in K(x) shows h ⊆ H. But then the contradiction h ∈ g∈G H = 1 follows. Hence, |P | = ps, |C| = pr−s − 1, and 1 ind(0) ≥ n − o(I) + n − oI(1) = n − 2. |C|

The remaining assertions follow directly from the genus-0 condition.

Now we show that an aﬃne polynomial of degree p2 is an AGL-polynomial of the form (5.1) if it fulﬁlls the conclusion of Proposition 5.8.

Proposition 5.9 Let f ∈ K[X] be a monic aﬃne polynomial of degree p2. Let x be a zero of f(X) − t and denote by 0 resp. P0 the zero place of K(t) resp. K(x). Suppose 0 decomposes in K(x) in the form

p2−p 0 = P0 · P1 ··· Pp with pairwise diﬀerent places Pi ∈ P K(x) .

Moreover assume that ind(0) = p2 −2 holds. Then f = Xp2 +aXp2−1 +cXp2−p with a, c ∈ K] and both arithmetic and geometric monodromy group of f are equal to AGL(2, p). Proof. Corollary 3.16 gives ind(∞) ≥ p2. The genus-0 condition shows immediately that no other ﬁnite place of K(t) ramiﬁes. We obtain

2 f(X) = Xp −p · g(X) with a polynomial g ∈ K[X] of degree p, g(0) 6= 0.

Since 0 is the unique ﬁnite place ramifying in K(x)|K(t), the polynomial f(X)−c is separable for all c ∈ K×. This shows that 0 ∈ K is the only root of f 0. The derivative of f is given by f 0(X) = Xp2−pg0(X). Thus, g0 is a monomial. We obtain

g(X) = Xp + aXb + c with 0 < b < p and ac 6= 0.

As Pi|0 is unramiﬁed for i ∈ {1, . . . , p}, the degree of the diﬀerent d(Pi|0) vanishes; by our 2 assumption d(P0|0) = p − 2. Stichtenoth [47, III.5.10(a)] gives

2 0 p2−p+b−1 2 p − 2 = d(P0|0) ≤ vP0 f (x) = vP0 abx = p − p + b − 1; hence, b = p − 1. The claim follows.

58 The results of this chapter allow us to give some properties of the ﬁxed ﬁeld E:

Remark 5.10 The genus g(E) is not bounded. p2 p2−1 p2−p Denote by Kp an algebraic closure of Fp and set fp(X) := X + X + X − t. Then 1 3 2 Gal fp|Kp(t) = AGL(2, p); Riemann-Hurwitz shows g(E) = 2 p − 3p + 4 . This proves g(E) → ∞ for p → ∞. pr pr−1 pr−p Let r ≥ 2 be an integer and set gp(X) := X + X + X . Then Gal gp|Kp(t) = AGL(r, p). Again, by Riemann-Hurwitz g(E) → ∞ for r → ∞. ?

Remark 5.11 Even if G is solvable, E need not be a rational function ﬁeld. 9 8 6 Let K denote an algebraic closure of F3 and deﬁne f(X) := X + X + X − t. Then Galf|K(t) =∼ AGL(2, 3) is solvable with g(E) = 2. ?

59 5. AGL as a monodromy group

60 6. Aﬃne polynomials of degree p2

In this chapter we classify all aﬃne polynomials of degree p2 with primitive arithmetic monodromy group. Our result will be

Theorem 6.1 With the notation from page 23 suppose deg f = p2. Then either E|K(t) is tame and f fulﬁlls the conclusion of Theorem 4.2 or one of the following cases holds: ∼ ∼ • H = GL(2, p), I∞ = N o Cn−1 with I∞(2) = 1 and ind(∞) = n, Ip = Cp o Cp−1 with Ip(2) = 1, ind(p) = n − 2, and o(Ip) = p + 1. This case is realized for example by the AGL-polynomial Xp2 + aXp2−1 + bXp2−p, cf. chapter 5.

0 • SL(2, p) ≤ H and Fp is a cyclic p -group.

If E|K(t) is wild, then f is not exceptional.

We know from Lemma 4.1 that E|K(t) being a tame extension implies g = 0. Hence, this case only gives polynomials satisfying the conclusion of Theorem 4.2. Therefore we will suppose E|K(t) to be wild from now on. We state a ﬁrst observation

Lemma 6.2 Suppose f has degree pr with r being a prime. Assume further that f is functionally indecomposable over k but decomposable over K. Then p - [E : K(t)]. In particular, the extension E|K(t) is tame.

Proof. As U is irreducible on N, the H-module N is semisimple by Cliﬀord. As H acts Lr reducibly, we can write N = i=1 Ni where the Ni are irreducible H-submodules of N of 0 order p. Since the automorphism group of Ni is abelian, the commutator subgroup H of H 0 lies in the kernel of the action of H on Ni. Thus, H is a subgroup of the kernel of H on N; hence, H0 = 1 and H is abelian. Let P be a p-Sylow subgroup of H. Then P is characteristic in H and, thus, normal in U. Proposition 2.8 gives P = 1; hence, p - [E : K(t)].

The above lemma allows us to reduce our classiﬁcation to polynomials that are indecomposable over K, or, equivalently, groups H ≤ GL(2, p) acting irreducibly. These groups fulﬁll

Proposition 6.3 Suppose H is an irreducible subgroup of GL(2, p). If p divides |H|, then SL(2, p) ≤ H. ∼ Proof. Let P = Cp denote a p-Sylow subgroup of GL(2, p). As p | |H|, we may assume P ≤ H. Proposition 2.8 shows that P is not normal in H. Hence, there exists h ∈ H with P 6= P h. [32, 8.6.7] gives hP,P hi = SL(2, p) ≤ H.

The next lemma states some well-known properties of GL(2, p):

61 6. Aﬃne polynomials of degree p2

Lemma 6.4 (1) Let P denote a p-Sylow subgroup of GL(2, p). Then the order of a p- regular element in NGL(2,p)(P ) divides p − 1.

(2) Let g ∈ SL(2, p) ≤ H be p-regular. Then g ﬁxes only 0 ∈ N.

6.1. Cases in odd characteristic p 6= 2

By Proposition 6.3 we may assume H = SL(2, p) o C where C is a cyclic group of order r|p − 1. Set Z := Fix N o SL(2, p) . Then Z|K(t) is a tame Galois extension of degree r. For the following s and t denote integers relatively prime to p. Two finite places ramify ∼ ∼ Suppose p and q ramify. By Lemma 3.21 we get Fp = Fq = C2. Since an involution in GL(2, p) fixes either exactly one point or exactly p points, ind(p) and ind(q) can only have the values 1 p2 − 1 1 p(p − 1) p2 − (p2 + 1) = or p2 − (p2 + p) = . 2 2 2 2 As E|K(t) is wild, ∞ ramifies in this extension with index ps. Lemma 6.4 shows s | p − 1; thus, 1 ind(∞) ≥ (p2 − 1) 1 + = p(p + 1). p − 1 We obtain a violation of the genus-0 condition. Only one finite place ramifies By Riemann-Hurwitz Z is a rational function field. The classification of Valentini and Madan [50] shows that Z|K(t) has exactly two branch points, both ramifying totally. Thus, r divides |F∞| and |Fp|.

∼ Suppose p ramiﬁes wildly. We have Fp = Cp o Crt with rt | p − 1. Set S := Fp ∩ SL(2, p). Then S has order pt and by Lemma 6.4 Ip contains (pt − p) elements that ﬁx at most one point. Thus,

1 1 1 ind(p) ≥ p2 − p2 + (p − 1)p + pt − p +(prt − pt)p + p2 − p2 + (p − 1)p . prt | {z } rt p induced by S

If ∞ ramiﬁes wildly in the extension E|K(t), then ind(∞) ≥ p(p + 1). This induces a contradiction to the genus-0 condition. ∼ 2 1 Hence, F∞ = Crs with ind(∞) ≥ (p − 1) 1 + rs . We obtain

(p − 1)t + pt + s(p + t − 3 − rt) ind(∞) + ind(p) − 2p2 + 2 ≥ . rst Since rt | p −1, the genus-0 condition is violated for t 6= 1. Thus, suppose t = 1. The genus-0 ∼ condition forces r = p − 1 and s = p + 1. This gives H = GL(2, p), Ip(2) = I∞(2) = 1, and ] 2 2 | Fix(g)| = p for all g ∈ Ip. It follows ind(∞) = p , ind(p) = p − 2, and o(Ip) = p + 1. There exist polynomials f that realize this case, cf. chapter 5. As H is transitive on N ], such an f cannot be exceptional.

62 6.2. Cases in even characteristic p = 2

If p ramifies tamely, then our estimations for the degree of the different of K(x)|K(t) are too weak to induce contradictions to the genus-0 condition. Hence, we only obtain SL(2, p) ≤ H 0 with Fp being a cyclic p -group. No finite ramification A discussion of the extension Z|K(t) shows r = 1. Hence, H =∼ SL(2, p) and ∞ ramifies with ∼ F∞ = Cp oCs where s | p−1. Set a ∈ N the well-defined integer with F∞(a) > F∞(a+1) = 1. Riemann-Hurwitz states |H| 2g(E) − 2 = −2|H| + ps − 1 + a · (p − 1). ps

p2−1+s(p3−p−2) As g(E) ≥ 0, we obtain a ≥ (1+p)(p−1)2 . Section 3.2.1 shows 1 1 ind(∞) ≥ (p2 − 1) 1 + + (a − 1)(p2 − p); s s this gives a contradiction to the genus-0 condition.

6.2. Cases in even characteristic p = 2 ∼ Proposition 6.3 gives H = GL(2, 2) = SL(2, 2) = S3. Denote by C the normal cyclic subgroup of H of order 3. Set Z := Fix(N o C). Then E|Z resp. Z|K(t) is Galois of degree 3 resp. 2. As all subgroups of H of order 2 are selfnormalizing, ramification groups in the extension E|K(t) are either isomorphic to C3 or to C2. Only one finite place ramifies ∼ ∼ By Corollary 3.16 F∞ = C3 and Ip = C2. Van der Waerden [51] shows that p decomposes in K(x) in the form

2 p = P · P1 · P2 with pairwise diﬀerent places of K(x).

Proposition 5.9 proves that the extension L|K(t) comes from a polynomial belonging to the class of AGL-polynomials described in chapter 5. No ﬁnite ramiﬁcation ∼ As E|K(t) is wild, we get F∞ = C2. By Section 3.2.1 F∞(2) = 1; but this gives a contradiction to the Riemann-Hurwitz genus formula.

63 6. Aﬃne polynomials of degree p2

64 7. Exceptional polynomials of degree p3

In this chapter we classify all exceptional polynomials of degree p3 with primitive arithmetic monodromy group. Our result will be

Theorem 7.1 With the notation from page 23 suppose deg f = p3. If f is exceptional, then the extension E|K(t) is tame. In particular, g = 0 and Theorem 4.2 lists all possibilities for f. We will suppose the extension E|K(t) to be wild from now on. Lemma 6.2 gives the irreducibility of H on N. We start with the discussion of the case p > 3.

7.1. The cases in characteristic p > 3

Suppose p > 3. We ﬁrst classify all irreducible subgroups H ≤ GL(3, p) in question.

Lemma 7.2 Suppose H ≤ GL(3, p) with p | |H|. Deﬁne S := H ∩ SL(3, p) and Z := ZSL(3, p). Set S := SZ/Z ≤ PSL(3, p) the image of S in PSL(3, p). Then:

(1) H acts irreducibly on N ⇐⇒ S acts irreducibly on N.

(2) If S is irreducible on N, then S =∼ PSL(3, p), S =∼ PSL(2, p), or S =∼ PGL(2, p).

(3) If H is irreducible on N but not transitive on N ], then

0 H0 ≤ H ≤ H0 × Z GL(3, p)

where H0 is isomorphic to PGL(2, p) and acts irreducibly on N. H0 is conjugate to the image of PGL(2, p) under the mapping (4.4) on page 39.

Proof.

(1) S is a subgroup of H; if S is irreducible, then H is a fortiori irreducible. Suppose H is irreducible and S acts reducibly. Let 1 6= P ≤ H be a p-Sylow subgroup of H. As H/S ≤ Cp−1, the group P ≤ S is also a p-Sylow subgroup of S. Thus, we obtain the same contradiction as in the proof of Lemma 6.2.

(2) We use Bloom’s [2] classiﬁcation of the subgroups of PSL(3, p). First assume that S does not have any nontrivial elementary abelian normal subgroup. Then [2, Thm. 1.1] and the condition p | |S| give S =∼ PSL(3, p), S =∼ PSL(2, p), or S =∼ PGL(3, p). We show in part (3) of this proof that these cases really induce an irreducible group S. Now assume S has a nontrivial normal elementary abelian subgroup; this case is dealt with in [2, Theorem 7.1]. The cases (1), (2), (4), and (5) of this theorem cannot occur for S would be a p0-group. In the remaining case (3) the group S contains a normal

65 7. Exceptional polynomials of degree p3

elementary abelian p-subgroup P . By Proposition 2.8 P = 1. Bloom [2] shows that S – up to conjugacy and/or the inverse-transpose isomorphism – embeds into the subgroup ∗ 0 0 of GL(3, p) consisting of elements of the form ∗ ∗ ∗ . This shows that S does not act ∗ ∗ ∗ irreducibly. (3) S =∼ PSL(3, p) enforces S = SL(3, p). But then S would be 2-transitive, a contradiction to our assumption. In the remaining cases S contains a normal subgroup isomorphic to PSL(2, p). Bloom [2, Lemma 6.3] shows that the commutator subgroup S0 of S is conjugate to the image of PSL(2, p) under the mapping ψ on page 37. In particular we obtain the irreducibility of S. 0 0 As S is characteristic in H, we see that H is a subgroup of NGL(3,p)(S ). Bloom proves 0 ∼ 0 0 that NSL(3,p)(S ) = H0 × Z where H0 = PGL(2, p) is irreducible with H0 = S . This shows that H0 is conjugate to the image of PGL(2, p) under the mapping (4.4) on 0 page 39. Certainly H0 × Z GL(3, p) ≤ NGL(3,p)(H0). Some calculation shows that 0 0 ∼ ∼ CGL(3,p)(H0) = Z GL(3, p) . As Aut(H0) = Aut PSL(2, p) = PGL(2, p), the N/C theorem gives 0 H0 × Z GL(3, p) = NGL(3,p)(H0).

Remark 7.3 The assumption p | |H| in statement (1) of the previous lemma is necessary. Consider for instance the group D 0 1 0 E H := 3 5 3 ≤ GL(3, 7). 1 0 2 ∼ H is isomorphic to C9 and acts irreducibly. But S := H ∩ SL(3, 7) = Z(SL(3, 7)) = C3 is reducible. ?

As we are interested in exceptional polynomials f, H is primitive on N but may not be transitive on N ]. Hence, all possibilities for H are given by Lemma 7.2 (3). As a consequence we get

0 0 Lemma 7.4 (1) The fixed field Z := Fix(NoH ) is rational. Either H/H is isomorphic to the Klein four-group; then exactly three places of K(t) ramify in the extension Z|K(t), each one with index r = 2. Or H/H0 is cyclic of order r | p − 1; then exactly two places of K(t) ramify in the extension Z|K(t), each one with index r. (2) Let h ∈ H be of order > 2. Then h has at most p fixed points.

(3) Let I be a subgroup of H being isomorphic to Cp o Cs. If gcd(s, p) = 1, then s | p − 1. Proof. (1) Lemma 7.2 shows that H/H0 is a p0-group. Thus, the extension Z|K(t) is tame with g(Z) = 0. As K is algebraically closed, it follows that Z is a rational function field. The ramification data and the group structure of H/H0 can be obtained directly from the classification of Valentini and Madan [50]. 0 H/H embeds into C2 × Cp−1; thus, every cyclic subgroup of C2 × Cp−1 has order a divisor of p − 1. This shows r | p − 1.

66 7.1. The cases in characteristic p > 3

(2) With the notation from Lemma 7.2 write h = gz with g ∈ H0 and z ∈ Z GL(3, p) . 2 2 2 2 0 2 2 Then h = g z 6= 1 and g ∈ H0. If g = 1, then h and, hence, h do not have any nontrivial ﬁxed point. Thus, suppose g2 6= 1. Then either g2 has order p; here the assertion follows from equation 4.3 on page 37. Or g2 is p-regular; but then the representation ψ on page 37 shows that g2 has three pairwise diﬀerent eigenvalues. Hence, the assertion is valid in this case, too. ∼ (3) We use the notation from Lemma 7.2. Suppose H = H0 × Z GL(3, p) . Let P be a p- Sylow subgroup of H. Then P ≤ H0 and NH (P ) = NH0 (P )×Z GL(3, p) . Huppert [25, ∼ ∼ II 7.1] shows NH0 (P ) = CpoCp−1. As also Z GL(3, p) = Cp−1, every p-regular element of NH (P ) has order a divisor of p − 1. This is the claim.

For the following s and t denote integers relatively prime to p. Two finite places ramify Both p and q ramify tamely with index 2. Therefore we obtain 1 ind(p) + ind(q) ≥ 2 · p3 − (p3 + p2) = p3 − p2. 2 3 1 Lemma 7.4 gives ind(∞) ≥ (p −1) 1+ p−1 . These estimations violate the genus-0 condition. No finite ramification 0 ∼ Lemma 7.4 yields H = H = PSL(2, p). Suppose |F∞| = ps. Let a ∈ N be the well-defined integer with F∞(a) > F∞(a + 1) = 1. Riemann-Hurwitz states |H| 2g(E) − 2 = −2|H| + ps − 1 + a(p − 1). ps As g(E) ≥ 0, we obtain p2 − 1 + s(p3 − p − 4) a ≥ . (p + 1)(p − 1)2 Section 3.2.1 shows 1 1 ind(∞) ≥ (p3 − 1) 1 + + (a − 1)(p3 − p); s s this gives a contradiction to the genus-0 condition. Exactly one finite place ramifies As at most two places ramify in Z|K(t), Lemma 7.4 shows that Gal(Z|K(t)) is cyclic of order r | p − 1 and both ∞ and p ramify totally in this extension.

∼ Suppose ﬁrst that p ramiﬁes wildly. Then Fp = Cp o Crt with rt | p − 1. Lemma 4.8 shows that Ip contains at most p involutions. Thus, 1 1 1 ind(p) ≥ p3 − p3 + p · p2 + (prt − p − 1)p + p3 − p3 + (p − 1)p . prt rt p

0 3 1 The p -part of the ramiﬁcation index of ∞ is given by rs. We have ind(∞) ≥ (p −1) 1+ rs . This yields t(p3 − 1) + s(p − 1)p(p − 2) − rt − 2 ind(∞) + ind(p) − 2p3 + 2 ≥ . rst

67 7. Exceptional polynomials of degree p3

As p(p − 2) − rt − 2 ≥ p(p − 3) − 1 > 0, this case is impossible.

∼ ∼ Now suppose that p ramifies tamely. Then Fp = Crt and F∞ = Cp o Crs with rs | p − 1. Let a denote the well-defined integer with F∞(a) > F∞(a + 1) = 1. For the following we use additionally the notation from section 3.2.1. ∼ ∼ We show first that I∞(ua) = Cp is impossible. Suppose I∞(ua) = Cp. Maus [37] gives I∞(ua) E I∞(1); thus, ∼ NI∞(1) I∞(ua) /CI∞(1) I∞(ua) ,→ Aut I∞(ua) = Cp−1.

0 As Aut(I∞(ua)) is a p -group, it follows I∞(1) = NI∞(1) I∞(ua) = CI∞(1) I∞(ua) .

3 The total ramiﬁcation of ∞ in K(x)|K(t) shows p | |I∞|. As a p-Sylow subgroup of H is 2 ] cyclic of order p, it follows p | |J∞(1)|. As I∞(ua) 6≤ N, there exist n ∈ N and h ∈ H such that I∞(ua) = hnhi. Moreover h cannot be an involution; hence, h ﬁxes at most p points. But for all m ∈ J∞(1)

nh · m = m · nh ⇐⇒ nhm = mnh = nmh ⇐⇒ hm = mh ⇐⇒ m = mh.

But this is impossible. 2 1 3 2 1 1 2 Hence, |I∞(ua)| ≥ p and o I∞(ua) ≤ p2 p + (p − 1)p = 2p − p . Since 2p − p < p , Corollary 3.17 gives o(I∞(ua)) ≤ p. Due to our classiﬁcation in chapter 4 we may assume the genus of E to be ≥ 3. The above estimations together with the estimation for a coming from the Riemann-Hurwitz genus formula succeed in disproving the genus-0 condition.

7.2. The cases in characteristic p ∈ {2, 3}

In this section we use the computational algebra system MAGMA [3] to classify all pairs (A, G) being exceptional with A a primitive group of degree 8 or 27. We obtain only one pair in characteristic 3 with H being elementary abelian of order 4. Hence, even in this case the extension E|K(t) is tame.

68 8. Exceptional polynomials of degree pr, r an odd prime, with 2-transitive group A

In this chapter we classify all exceptional polynomials of degree pr, r an odd prime, such that the arithmetic monodromy group of f is 2-transitive. Our result will be

Theorem 8.1 With the notation from page 23 suppose deg f = pr. Assume the arithmetic monodromy group of f is 2-transitive. If f is exceptional, then the extension E|K(t) is tame. In particular, g = 0 and Theorem 4.2 lists all possibilities for f. The 2-transitivity of A on N gives the transitivity of U on N ]. The following lemma is based on Hering’s classiﬁcation of transitive subgroups of GL(r, p). A complete treatment of these groups can be found in Liebeck [36, Appendix 1].

Lemma 8.2 N can be considered a one-dimensional Fpr vector space. U acts on N as a subgroup of ΓL(1, pr). Proof. r being an odd prime enforces either U ≤ ΓL(1, pr) or SL(r, p) ≤ U. Suppose the latter case holds. Huppert [25, II 6.10] shows the perfectness of SL(r, p); thus, we have SL(r, p) ≤ H. However, the transitivity of SL(r, p) on N ] is a contradiction to the exceptionality of f.

We assume the extension E|K(t) to be wild from now on. As the order of ΓL(1, pr) is |ΓL(1, pr)| = (pr−1)·r, this immediately yields p = r. The next lemma gives some estimations for the number of ﬁxed points of elements of A.

Lemma 8.3 (1) Let 1 6= g ∈ A ≤ AΓL(1, pp). Then g ﬁxes at most p elements. If g is p-regular, then it ﬁxes exactly one point.

p (2) Let P be a p-Sylow subgroup of ΓL(1, p ). Then the normalizer NΓL(1,pp)(P ) is cyclic of order p(p − 1). Proof.

(1) As we are interested in the maximal number of fixed points of an element 1 6= g ∈ A, ∼ Lemma 3.12 allows us to assume g ∈ U. Lemma 8.2 gives the identification N = Fpp . × Thus, the action of g on N is given by a tuple (a, b) with 0 ≤ a < p and b ∈ Fpp such that a ng = np · b for all n ∈ N. Suppose n ∈ N ] is fixed by g. Then n fulfills npa−1 = b−1. If m ∈ N ] is also fixed by g, ] pa−1 × then we have m = d · n with d ∈ N and d = 1. Thus, d ∈ Fp and Fix(g) = n · Fp. Suppose g is p-regular. A simple induction shows that the action of gp is given by (1, β) with β 6= 1. It follows that the unique fixed point of gp and, thus, of g is the zero element of N. The assertion is due to Corollary 3.14.

69 8. Exceptional polynomials of degree pr, r an odd prime, with 2-transitive group A

(2) Since P is cyclic of order p, we may assume P to be generated by an element g ∈ p ΓL(1, p ) whose action is given by (1, 1). A simple calculation shows that NΓL(1,pp)(P ) × is generated by g and the (0, β)-element h where β denotes a primitive element of Fp . As g and h commute, we get ∼ ∼ NΓL(1,pp)(P ) = hg, hi = Cp × Cp−1 = Cp(p−1).

Now we discuss the ramification behavior of the extension E|K(t) in detail. As in the previous two chapters s and t denote integers relatively prime to p. Two finite places ramify Lemma 3.21 shows that both p and q ramify tamely with index 2. By Lemma 8.3 we have 1 ind(p) + ind(q) ≥ 2 n − n + 1 = n − 1. 2 Since ind(∞) ≥ n, this is a contradiction to the genus-0 condition. One finite place ramifies Suppose p ramifies wildly. Then Ip embeds into the normalizer of a p-Sylow subgroup of p ∼ ΓL(1, p ); hence, Ip = Cp × Ct with t | p − 1. If g ∈ Ip does not have order 1 or p, the p-th power of g is p-regular. Thus, by Lemma 8.3 only elements of order 1 or p can have more than one fixed point. This shows 1 1 1 ind(p) ≥ n − n + (p − 1)p + (pt − p) + n − n + (p − 1)p . pt t p Together with ind(∞) ≥ n we get n(p − 2) + p(t + 3 − 2p) ind(∞) + ind(p) − 2n + 2 ≥ ; pt as p > 2 and n = pp > 2p2, this case cannot occur.

t−1 Suppose p ramifies tamely with index t. Then ind(p) ≥ (n − 1) t . The group F∞ is cyclic of order ps. Let a ∈ N denote the well-defined integer with F∞(a) > F∞(a + 1) = 1. Then Riemann-Hurwitz gives |H| |H| 2g(E) − 2 = −2|H| + (t − 1) + ps − 1 + a(p − 1). t ps Due to the classification in chapter 4 we may assume g ≥ 3. This gives 4pst + |H|(ps + t) a ≥ . |H|t(p − 1) Together with the estimation of section 3.2.1 for ind(∞) a a violation of the genus-0 condition results. Only ∞ ramifies We use the same idea as above. We obtain ps(|H| + 4) + |H| a ≥ . |H|(p − 1) This again induces a contradiction to the genus-0 condition.

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